Solution of trigonometrical equations Questions in English

(C) Given equation: $2\sin^3\alpha - 7\sin^2\alpha + 7\sin\alpha - 2 = 0$.
Let $x = \sin\alpha$. Then $2x^3 - 7x^2 + 7x - 2 = 0$.
By testing roots,$x = 1$ is a root since $2(1)^3 - 7(1)^2 + 7(1) - 2 = 0$.
Dividing by $(x - 1)$,we get $(x - 1)(2x^2 - 5x + 2) = 0$.
Factoring the quadratic: $(x - 1)(2x - 1)(x - 2) = 0$.
So,$\sin\alpha = 1$,$\sin\alpha = \frac{1}{2}$,or $\sin\alpha = 2$.
Since $-1 \leq \sin\alpha \leq 1$,$\sin\alpha = 2$ is impossible.
For $\sin\alpha = 1$ in $[0, 2\pi]$,$\alpha = \frac{\pi}{2}$ ($1$ value).
For $\sin\alpha = \frac{1}{2}$ in $[0, 2\pi]$,$\alpha = \frac{\pi}{6}, \frac{5\pi}{6}$ ($2$ values).
Total number of values = $1 + 2 = 3$.

(A) Given equation: $\sin 2x - 2 \cos x + 4 \sin x = 4$
Using $\sin 2x = 2 \sin x \cos x$,we get:
$2 \sin x \cos x - 2 \cos x + 4 \sin x - 4 = 0$
Factor by grouping:
$2 \cos x (\sin x - 1) + 4 (\sin x - 1) = 0$
$(2 \cos x + 4)(\sin x - 1) = 0$
$2(\cos x + 2)(\sin x - 1) = 0$
Since $\cos x + 2 \neq 0$ for any real $x$ (as $-1 \le \cos x \le 1$),we must have:
$\sin x = 1$
In the interval $[0, 5\pi]$,the values of $x$ for which $\sin x = 1$ are:
$x = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}$
Thus,there are $3$ solutions.

(A) Given equation: $\sin x - \sin 2x + \sin 3x = 0$
Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$(\sin 3x + \sin x) - \sin 2x = 0$
$2 \sin 2x \cos x - \sin 2x = 0$
$\sin 2x (2 \cos x - 1) = 0$
This implies $\sin 2x = 0$ or $\cos x = \frac{1}{2}$.
For $\sin 2x = 0$,$2x = n\pi$,so $x = \frac{n\pi}{2}$. Given $0 \le x < \frac{\pi}{2}$,the only solution is $x = 0$.
For $\cos x = \frac{1}{2}$,$x = \frac{\pi}{3}$ (since $0 \le x < \frac{\pi}{2}$).
The values of $x$ are $0$ and $\frac{\pi}{3}$.
Thus,the number of values is $2$.

(C) Given equation: $\sin^2 2\theta + \cos^4 2\theta = \frac{3}{4}$
Using $\sin^2 2\theta = 1 - \cos^2 2\theta$,we get:
$1 - \cos^2 2\theta + \cos^4 2\theta = \frac{3}{4}$
Let $t = \cos^2 2\theta$. Then $t^2 - t + 1 = \frac{3}{4} \Rightarrow t^2 - t + \frac{1}{4} = 0$
$(t - \frac{1}{2})^2 = 0 \Rightarrow t = \frac{1}{2}$
So,$\cos^2 2\theta = \frac{1}{2} \Rightarrow 2\cos^2 2\theta - 1 = 0$
Using the identity $\cos 4\theta = 2\cos^2 2\theta - 1$,we have $\cos 4\theta = 0$
For $\theta \in (0, \frac{\pi}{2})$,$4\theta \in (0, 2\pi)$.
$\cos 4\theta = 0 \Rightarrow 4\theta = \frac{\pi}{2}, \frac{3\pi}{2}$
$\theta = \frac{\pi}{8}, \frac{3\pi}{8}$
Sum of values = $\frac{\pi}{8} + \frac{3\pi}{8} = \frac{4\pi}{8} = \frac{\pi}{2}$

(B) Given equation: $2\cos^2 \theta + 3\sin \theta = 0$
Using $\cos^2 \theta = 1 - \sin^2 \theta$,we get:
$2(1 - \sin^2 \theta) + 3\sin \theta = 0$
$2 - 2\sin^2 \theta + 3\sin \theta = 0$
$2\sin^2 \theta - 3\sin \theta - 2 = 0$
$(2\sin \theta + 1)(\sin \theta - 2) = 0$
Since $\sin \theta = 2$ is impossible,we have $\sin \theta = -\frac{1}{2}$.
For $\theta \in [-2\pi, 2\pi]$,the solutions for $\sin \theta = -\frac{1}{2}$ are:
$\theta = -\frac{\pi}{6}, -\frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$.
Sum of elements = $(-\frac{\pi}{6}) + (-\frac{5\pi}{6}) + (\frac{7\pi}{6}) + (\frac{11\pi}{6}) = \frac{-6\pi + 18\pi}{6} = \frac{12\pi}{6} = 2\pi$.

(C) Given the equation $1 + \sin^4 x = \cos^2 3x$.
Since $\sin^4 x \ge 0$,the left side $1 + \sin^4 x \ge 1$.
Since $\cos^2 3x \le 1$,the right side $\cos^2 3x \le 1$.
For the equality to hold,we must have $1 + \sin^4 x = 1$ and $\cos^2 3x = 1$.
$1 + \sin^4 x = 1 \implies \sin^4 x = 0 \implies \sin x = 0 \implies x = n\pi$ for $n \in \mathbb{Z}$.
Now check $\cos^2 3x = 1$ for $x = n\pi$:
$\cos^2(3n\pi) = (\pm 1)^2 = 1$. This is always true for any integer $n$.
We need $x \in [-\frac{5\pi}{2}, \frac{5\pi}{2}]$,which is $[-2.5\pi, 2.5\pi]$.
The possible values for $x = n\pi$ in this interval are:
$n = -2 \implies x = -2\pi$
$n = -1 \implies x = -\pi$
$n = 0 \implies x = 0$
$n = 1 \implies x = \pi$
$n = 2 \implies x = 2\pi$
The solutions are $\{-2\pi, -\pi, 0, \pi, 2\pi\}$.
There are $5$ solutions.

(A) Given equation: $\log _{1 / 2}|\sin x|=2-\log _{1 / 2}|\cos x|$ for $x \in [0, 2\pi]$.
Rearranging the terms,we get: $\log _{1 / 2}|\sin x| + \log _{1 / 2}|\cos x| = 2$.
Using the property $\log_a m + \log_a n = \log_a (mn)$,we have: $\log _{1 / 2}(|\sin x \cos x|) = 2$.
Converting to exponential form: $|\sin x \cos x| = (1/2)^2 = 1/4$.
Multiplying by $2$ on both sides: $|2 \sin x \cos x| = 2 \times (1/4) = 1/2$.
Thus,$|\sin 2x| = 1/2$.
In the interval $x \in [0, 2\pi]$,the angle $2x$ lies in the interval $[0, 4\pi]$.
For $|\sin \theta| = 1/2$,there are $4$ solutions in each interval of length $2\pi$ (i.e.,$[0, 2\pi]$).
Since the interval for $2x$ is $[0, 4\pi]$,which covers two full periods of the sine function,the total number of solutions is $4 \times 2 = 8$.

(A) We know that $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$.
Since $\sin x$ is positive in the first and second quadrants,the other solution is in the second quadrant.
$\sin \left( \pi - \frac{\pi}{3} \right) = \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}$.
Therefore,the principal solutions are $x = \frac{\pi}{3}$ and $x = \frac{2\pi}{3}$.

(A) We know that $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}.$
Since $\tan x$ is negative in the second and fourth quadrants,we have:
$\tan x = \tan(\pi - \frac{\pi}{6}) = \tan \frac{5 \pi}{6} = -\frac{1}{\sqrt{3}}$
and
$\tan x = \tan(2 \pi - \frac{\pi}{6}) = \tan \frac{11 \pi}{6} = -\frac{1}{\sqrt{3}}.$
Thus,the principal solutions are $\frac{5 \pi}{6}$ and $\frac{11 \pi}{6}.$

(A) Given $\sin x = -\frac{\sqrt{3}}{2}$.
Since $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$,we have $\sin x = -\sin \frac{\pi}{3}$.
Using the identity $\sin(\pi + \theta) = -\sin \theta$,we get $\sin x = \sin(\pi + \frac{\pi}{3}) = \sin \frac{4\pi}{3}$.
The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$,where $n \in \mathbb{Z}$.
Substituting $\alpha = \frac{4\pi}{3}$,we get $x = n\pi + (-1)^n \frac{4\pi}{3}$,where $n \in \mathbb{Z}$.

(A) We have,$\cos x = \frac{1}{2} = \cos \frac{\pi}{3}$.
The general solution for $\cos x = \cos \alpha$ is given by $x = 2n\pi \pm \alpha$,where $n \in Z$.
Therefore,$x = 2n\pi \pm \frac{\pi}{3}$,where $n \in Z$.

(N/A) We have,$\tan 2x = -\cot \left(x + \frac{\pi}{3}\right)$
Since $-\cot \theta = \tan \left(\frac{\pi}{2} + \theta\right)$,we get:
$\tan 2x = \tan \left(\frac{\pi}{2} + x + \frac{\pi}{3}\right)$
$\tan 2x = \tan \left(x + \frac{5\pi}{6}\right)$
The general solution for $\tan \theta = \tan \alpha$ is $\theta = n\pi + \alpha$,where $n \in \mathbb{Z}$.
Therefore,$2x = n\pi + x + \frac{5\pi}{6}$
$x = n\pi + \frac{5\pi}{6}$,where $n \in \mathbb{Z}$.

The given equation is $\sin 6x + \sin 2x - \sin 4x = 0$.
Using the sum-to-product formula $\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$,we get:
$2 \sin 4x \cos 2x - \sin 4x = 0$.
Factoring out $\sin 4x$:
$\sin 4x (2 \cos 2x - 1) = 0$.
This implies either $\sin 4x = 0$ or $\cos 2x = \frac{1}{2}$.
For $\sin 4x = 0$,the general solution is $4x = n\pi$,which gives $x = \frac{n\pi}{4}$ for $n \in \mathbb{Z}$.
For $\cos 2x = \frac{1}{2} = \cos \frac{\pi}{3}$,the general solution is $2x = 2n\pi \pm \frac{\pi}{3}$,which gives $x = n\pi \pm \frac{\pi}{6}$ for $n \in \mathbb{Z}$.

(N/A) The given equation is $2 \cos^{2} x + 3 \sin x = 0$.
Using the identity $\cos^{2} x = 1 - \sin^{2} x$,we get:
$2(1 - \sin^{2} x) + 3 \sin x = 0$
$2 - 2 \sin^{2} x + 3 \sin x = 0$
$2 \sin^{2} x - 3 \sin x - 2 = 0$
Factoring the quadratic equation:
$(2 \sin x + 1)(\sin x - 2) = 0$
This gives two cases:
$1) \sin x = -\frac{1}{2}$
$2) \sin x = 2$
Since the range of $\sin x$ is $[-1, 1]$,$\sin x = 2$ is impossible.
For $\sin x = -\frac{1}{2}$,we know $\sin x = \sin(-\frac{\pi}{6})$.
The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^{n}\alpha$.
Thus,$x = n\pi + (-1)^{n}(-\frac{\pi}{6})$,where $n \in \mathbb{Z}$.

(A) Given equation: $\tan x = \sqrt{3}$.
We know that $\tan \frac{\pi}{3} = \sqrt{3}$.
Since $\tan x$ is positive in the first and third quadrants,the principal solutions are $x = \frac{\pi}{3}$ and $x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$.
For the general solution,we use the property that if $\tan x = \tan \alpha$,then $x = n\pi + \alpha$,where $n \in \mathbb{Z}$.
Here,$\alpha = \frac{\pi}{3}$.
Therefore,the general solution is $x = n\pi + \frac{\pi}{3}$,where $n \in \mathbb{Z}$.

(A) Given $\sec x = 2$.
Since $\sec x = \frac{1}{\cos x}$,we have $\cos x = \frac{1}{2}$.
We know that $\cos \frac{\pi}{3} = \frac{1}{2}$.
Since $\cos x$ is positive in the first and fourth quadrants,the principal solutions are $x = \frac{\pi}{3}$ and $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$.
For the general solution,if $\cos x = \cos \alpha$,then $x = 2n\pi \pm \alpha$,where $n \in \mathbb{Z}$.
Here $\alpha = \frac{\pi}{3}$,so the general solution is $x = 2n\pi \pm \frac{\pi}{3}$,where $n \in \mathbb{Z}$.

Given equation: $\cot x = -\sqrt{3}$.
We know that $\cot \frac{\pi}{6} = \sqrt{3}$.
Since $\cot x$ is negative in the second and fourth quadrants,we have:
$\cot (\pi - \frac{\pi}{6}) = -\cot \frac{\pi}{6} = -\sqrt{3} \Rightarrow \cot \frac{5\pi}{6} = -\sqrt{3}$.
$\cot (2\pi - \frac{\pi}{6}) = -\cot \frac{\pi}{6} = -\sqrt{3} \Rightarrow \cot \frac{11\pi}{6} = -\sqrt{3}$.
Thus,the principal solutions are $x = \frac{5\pi}{6}$ and $x = \frac{11\pi}{6}$.
For the general solution,we use the property that if $\cot x = \cot \alpha$,then $x = n\pi + \alpha$,where $n \in \mathbb{Z}$.
Taking the principal value $\alpha = \frac{5\pi}{6}$,the general solution is $x = n\pi + \frac{5\pi}{6}$,where $n \in \mathbb{Z}$.

(N/A) Given $\csc x = -2$.
We know that $\csc \frac{\pi}{6} = 2$.
Since $\csc x$ is negative in the third and fourth quadrants,we have:
$\csc(\pi + \frac{\pi}{6}) = -\csc \frac{\pi}{6} = -2$ and $\csc(2\pi - \frac{\pi}{6}) = -\csc \frac{\pi}{6} = -2$.
Thus,$\csc \frac{7\pi}{6} = -2$ and $\csc \frac{11\pi}{6} = -2$.
The principal solutions are $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$.
To find the general solution,we use $\csc x = \csc \frac{7\pi}{6}$,which implies $\sin x = \sin \frac{7\pi}{6}$.
The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$,where $n \in \mathbb{Z}$.
Substituting $\alpha = \frac{7\pi}{6}$,the general solution is $x = n\pi + (-1)^n \frac{7\pi}{6}$,where $n \in \mathbb{Z}$.

(A) Given equation: $\cos 4x = \cos 2x$
$\Rightarrow \cos 4x - \cos 2x = 0$
Using the formula $\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$:
$-2 \sin \left(\frac{4x+2x}{2}\right) \sin \left(\frac{4x-2x}{2}\right) = 0$
$-2 \sin 3x \sin x = 0$
$\sin 3x \sin x = 0$
This implies $\sin 3x = 0$ or $\sin x = 0$.
If $\sin 3x = 0$,then $3x = n\pi \Rightarrow x = \frac{n\pi}{3}$,where $n \in \mathbb{Z}$.
If $\sin x = 0$,then $x = n\pi$,where $n \in \mathbb{Z}$.
Combining these,the general solution is $x = \frac{n\pi}{3}$ or $x = n\pi$,where $n \in \mathbb{Z}$.

Given equation: $\cos 3x + \cos x - \cos 2x = 0$
Using the identity $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$:
$2 \cos \left( \frac{3x+x}{2} \right) \cos \left( \frac{3x-x}{2} \right) - \cos 2x = 0$
$2 \cos 2x \cos x - \cos 2x = 0$
$\cos 2x (2 \cos x - 1) = 0$
This gives two cases:
Case $1$: $\cos 2x = 0$ $\Rightarrow 2x = (2n+1) \frac{\pi}{2}$ $\Rightarrow x = (2n+1) \frac{\pi}{4}$,where $n \in \mathbb{Z}$.
Case $2$: $2 \cos x - 1 = 0$ $\Rightarrow \cos x = \frac{1}{2} = \cos \frac{\pi}{3}$ $\Rightarrow x = 2n\pi \pm \frac{\pi}{3}$,where $n \in \mathbb{Z}$.

(A) Given equation: $\sin 2x + \cos x = 0$
Using the identity $\sin 2x = 2 \sin x \cos x$,we get:
$2 \sin x \cos x + \cos x = 0$
Factoring out $\cos x$:
$\cos x (2 \sin x + 1) = 0$
This gives two cases:
Case $1$: $\cos x = 0$
The general solution for $\cos x = 0$ is $x = (2n + 1) \frac{\pi}{2}$,where $n \in \mathbb{Z}$.
Case $2$: $2 \sin x + 1 = 0$
$\sin x = -\frac{1}{2}$
Since $\sin \frac{\pi}{6} = \frac{1}{2}$,we have $\sin x = -\sin \frac{\pi}{6} = \sin(\pi + \frac{\pi}{6}) = \sin \frac{7\pi}{6}$.
The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$.
Thus,$x = n\pi + (-1)^n \frac{7\pi}{6}$,where $n \in \mathbb{Z}$.
Therefore,the general solution is $x = (2n + 1) \frac{\pi}{2}$ or $x = n\pi + (-1)^n \frac{7\pi}{6}$,where $n \in \mathbb{Z}$.

(A) Given equation: $\sec^{2} 2x = 1 - \tan 2x$
Using the identity $\sec^{2} \theta = 1 + \tan^{2} \theta$,we get:
$1 + \tan^{2} 2x = 1 - \tan 2x$
$\tan^{2} 2x + \tan 2x = 0$
$\tan 2x(\tan 2x + 1) = 0$
This gives two cases:
Case $1$: $\tan 2x = 0$
$2x = n\pi$,where $n \in \mathbb{Z}$
$x = \frac{n\pi}{2}, n \in \mathbb{Z}$
Case $2$: $\tan 2x = -1$
$\tan 2x = -\tan \frac{\pi}{4} = \tan(\pi - \frac{\pi}{4}) = \tan \frac{3\pi}{4}$
$2x = n\pi + \frac{3\pi}{4}$,where $n \in \mathbb{Z}$
$x = \frac{n\pi}{2} + \frac{3\pi}{8}, n \in \mathbb{Z}$
Thus,the general solution is $x = \frac{n\pi}{2}$ or $x = \frac{n\pi}{2} + \frac{3\pi}{8}, n \in \mathbb{Z}$.

Given equation: $\sin x + \sin 3x + \sin 5x = 0$
Rearranging the terms: $(\sin 5x + \sin x) + \sin 3x = 0$
Using the formula $\sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$:
$2 \sin \left(\frac{5x+x}{2}\right) \cos \left(\frac{5x-x}{2}\right) + \sin 3x = 0$
$2 \sin 3x \cos 2x + \sin 3x = 0$
Taking $\sin 3x$ as a common factor:
$\sin 3x (2 \cos 2x + 1) = 0$
This gives two cases:
Case $1$: $\sin 3x = 0$
$3x = n\pi$,where $n \in \mathbb{Z}$
$x = \frac{n\pi}{3}$,where $n \in \mathbb{Z}$
Case $2$: $2 \cos 2x + 1 = 0$
$\cos 2x = -\frac{1}{2}$
Since $\cos \frac{\pi}{3} = \frac{1}{2}$,we have $\cos 2x = -\cos \frac{\pi}{3} = \cos \left(\pi - \frac{\pi}{3}\right) = \cos \frac{2\pi}{3}$
$2x = 2n\pi \pm \frac{2\pi}{3}$,where $n \in \mathbb{Z}$
$x = n\pi \pm \frac{\pi}{3}$,where $n \in \mathbb{Z}$
Thus,the general solution is $x = \frac{n\pi}{3}$ or $x = n\pi \pm \frac{\pi}{3}$,where $n \in \mathbb{Z}$.

(A) The given equation is $|\cot x| = \cot x + \frac{1}{\sin x}$.
Case $1$: If $\cot x \ge 0$,then $|\cot x| = \cot x$. The equation becomes $\cot x = \cot x + \frac{1}{\sin x}$,which implies $\frac{1}{\sin x} = 0$. This has no solution.
Case $2$: If $\cot x < 0$,then $|\cot x| = -\cot x$. The equation becomes $-\cot x = \cot x + \frac{1}{\sin x}$,which simplifies to $2\cot x + \frac{1}{\sin x} = 0$.
Substituting $\cot x = \frac{\cos x}{\sin x}$,we get $2\frac{\cos x}{\sin x} + \frac{1}{\sin x} = 0$,so $\frac{2\cos x + 1}{\sin x} = 0$.
This implies $2\cos x + 1 = 0$,so $\cos x = -\frac{1}{2}$.
In the interval $[0, 2\pi]$,$\cos x = -\frac{1}{2}$ at $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$.
Check the condition $\cot x < 0$: For $x = \frac{2\pi}{3}$,$\cot(\frac{2\pi}{3}) = -\frac{1}{\sqrt{3}} < 0$ (Valid).
For $x = \frac{4\pi}{3}$,$\cot(\frac{4\pi}{3}) = \frac{1}{\sqrt{3}} > 0$ (Invalid,as it contradicts $\cot x < 0$).
Thus,there is only $1$ solution.

(A) To find the number of solutions for $x + 2 \tan x = \frac{\pi}{2}$ in the interval $[0, 2\pi]$,we rewrite the equation as:
$2 \tan x = \frac{\pi}{2} - x$
$\tan x = -\frac{1}{2}x + \frac{\pi}{4}$
We look for the intersection points of the graphs $y = \tan x$ and $y = -\frac{1}{2}x + \frac{\pi}{4}$ within the interval $[0, 2\pi]$.
$1$. In the interval $[0, \frac{\pi}{2})$,$\tan x$ increases from $0$ to $\infty$,while the line $y = -\frac{1}{2}x + \frac{\pi}{4}$ decreases from $\frac{\pi}{4} \approx 0.785$ to $0$. There is exactly $1$ intersection point.
$2$. In the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$,$\tan x$ increases from $-\infty$ to $\infty$,while the line decreases from $0$ to $-\frac{\pi}{2} \approx -1.57$. There is exactly $1$ intersection point.
$3$. In the interval $(\frac{3\pi}{2}, 2\pi]$,$\tan x$ increases from $-\infty$ to $0$,while the line decreases from $-\frac{\pi}{2}$ to $-\frac{3\pi}{4} \approx -2.35$. There is exactly $1$ intersection point.
Thus,there are a total of $3$ solutions.

(A) Given equation: $\sqrt{3} \cos^2 x = (\sqrt{3}-1) \cos x + 1$
Rearranging the terms: $\sqrt{3} \cos^2 x - \sqrt{3} \cos x + \cos x - 1 = 0$
Factorizing by grouping: $\sqrt{3} \cos x (\cos x - 1) + 1 (\cos x - 1) = 0$
$(\sqrt{3} \cos x + 1)(\cos x - 1) = 0$
This gives two cases:
Case $1$: $\cos x = 1 \Rightarrow x = 0$ (which is in the interval $[0, \frac{\pi}{2}]$)
Case $2$: $\cos x = -\frac{1}{\sqrt{3}}$. Since $\cos x$ is negative in the second quadrant and $x \in [0, \frac{\pi}{2}]$,there is no solution for this case in the given interval.
Thus,there is only $1$ solution,which is $x = 0$.

(C) Given equation is $\sin^{4} \theta + \cos^{4} \theta - \sin \theta \cos \theta = 0$.
We know that $\sin^{4} \theta + \cos^{4} \theta = (\sin^{2} \theta + \cos^{2} \theta)^{2} - 2\sin^{2} \theta \cos^{2} \theta = 1 - 2\sin^{2} \theta \cos^{2} \theta$.
Substituting this into the equation: $1 - 2\sin^{2} \theta \cos^{2} \theta - \sin \theta \cos \theta = 0$.
Using $2\sin \theta \cos \theta = \sin 2\theta$,we have $\sin^{2} \theta \cos^{2} \theta = \frac{\sin^{2} 2\theta}{4}$.
So,$1 - 2(\frac{\sin^{2} 2\theta}{4}) - \frac{\sin 2\theta}{2} = 0$.
Multiplying by $2$: $2 - \sin^{2} 2\theta - \sin 2\theta = 0$,which simplifies to $\sin^{2} 2\theta + \sin 2\theta - 2 = 0$.
Factoring gives $(\sin 2\theta + 2)(\sin 2\theta - 1) = 0$.
Since $\sin 2\theta$ cannot be $-2$,we have $\sin 2\theta = 1$.
For $\theta \in [0, 4\pi]$,$2\theta \in [0, 8\pi]$.
Thus,$2\theta = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, \frac{13\pi}{2}$.
$\theta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}$.
The sum $S = \frac{\pi + 5\pi + 9\pi + 13\pi}{4} = \frac{28\pi}{4} = 7\pi$.
Therefore,$\frac{8S}{\pi} = \frac{8(7\pi)}{\pi} = 56$.

(D) Given equation: $2 \cos x(4 \sin(\frac{\pi}{4}+x) \sin(\frac{\pi}{4}-x)-1)=1$
Using the identity $\sin(A+B)\sin(A-B) = \sin^2 A - \sin^2 B$,we get:
$2 \cos x(4(\sin^2(\frac{\pi}{4}) - \sin^2 x) - 1) = 1$
$2 \cos x(4(\frac{1}{2} - \sin^2 x) - 1) = 1$
$2 \cos x(2 - 4\sin^2 x - 1) = 1$
$2 \cos x(1 - 4\sin^2 x) = 1$
Since $1 - 4\sin^2 x = 1 - 4(1 - \cos^2 x) = 4\cos^2 x - 3$,the equation becomes:
$2 \cos x(4\cos^2 x - 3) = 1$
$8\cos^3 x - 6\cos x = 1$
$4\cos^3 x - 3\cos x = \frac{1}{2}$
Using the triple angle identity $\cos 3x = 4\cos^3 x - 3\cos x$,we have:
$\cos 3x = \frac{1}{2}$
Given $x \in [0, \pi]$,then $3x \in [0, 3\pi]$.
The solutions for $3x$ are $\frac{\pi}{3}, 2\pi - \frac{\pi}{3}, 2\pi + \frac{\pi}{3}$,which are $\frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}$.
Thus,$x = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}$.
The number of solutions $n = 3$.
The sum $S = \frac{\pi}{9} + \frac{5\pi}{9} + \frac{7\pi}{9} = \frac{13\pi}{9}$.
Wait,re-evaluating the sum: $\frac{1+5+7}{9}\pi = \frac{13\pi}{9}$. Checking options,there might be a typo in the provided options or the question. Given the structure,let's re-verify the sum. The solutions are $\frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}$. Sum is $\frac{13\pi}{9}$. If the question intended $x \in [0, 2\pi]$,the sum would be different. Based on the provided options,let's re-check the calculation. $4\cos^3 x - 3\cos x = 1/2 \implies \cos 3x = 1/2$. $3x = \pi/3, 5\pi/3, 7\pi/3$. $x = \pi/9, 5\pi/9, 7\pi/9$. Sum $= 13\pi/9$. None of the options match $13\pi/9$. If $n=3$,option $D$ is $(3, 5\pi/3)$. Let's re-read the graph. The graph shows $x = \pi/3, 5\pi/3, 7\pi/3$ for $\cos x = 1/2$. The equation derived is $\cos 3x = 1/2$. The solutions are $x = \pi/9, 5\pi/9, 7\pi/9$. The sum is $13\pi/9$. Given the options,there is likely a typo in the question's range or options. Assuming $n=3$,we select $D$ as the closest structure.

(A) Given $\sin^{7} x + \cos^{7} x = 1$ for $x \in [0, 4\pi]$.
Since $\sin^{2} x \leq 1$ and $\cos^{2} x \leq 1$,for $x \in [0, \pi/2]$,we have $\sin^{7} x \leq \sin^{2} x$ and $\cos^{7} x \leq \cos^{2} x$.
Adding these,$\sin^{7} x + \cos^{7} x \leq \sin^{2} x + \cos^{2} x = 1$.
The equality holds only when $\sin^{7} x = \sin^{2} x$ and $\cos^{7} x = \cos^{2} x$.
This implies $(\sin x = 0 \text{ or } \sin x = 1)$ and $(\cos x = 0 \text{ or } \cos x = 1)$.
Case $1$: $\sin x = 0 \implies x = 0, \pi, 2\pi, 3\pi, 4\pi$. Checking $\cos^{7} x = 1$,we get $x = 0, 2\pi, 4\pi$.
Case $2$: $\cos x = 0 \implies x = \pi/2, 3\pi/2, 5\pi/2, 7\pi/2$. Checking $\sin^{7} x = 1$,we get $x = \pi/2, 5\pi/2$.
Combining these,the solutions are $x \in \{0, \pi/2, 2\pi, 5\pi/2, 4\pi\}$.
Thus,there are $5$ solutions.

(D) Given equation: $(\sin x + \sin 4x) + (\sin 2x + \sin 3x) = 0$
Using the sum-to-product formula $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$:
$2 \sin \frac{5x}{2} \cos \frac{3x}{2} + 2 \sin \frac{5x}{2} \cos \frac{x}{2} = 0$
$2 \sin \frac{5x}{2} (\cos \frac{3x}{2} + \cos \frac{x}{2}) = 0$
Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$2 \sin \frac{5x}{2} (2 \cos x \cos \frac{x}{2}) = 0$
$4 \sin \frac{5x}{2} \cos x \cos \frac{x}{2} = 0$
Case $1$: $\sin \frac{5x}{2} = 0 \Rightarrow \frac{5x}{2} = n\pi \Rightarrow x = \frac{2n\pi}{5}$. For $x \in [0, 2\pi]$,$x \in \{0, \frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}, \frac{8\pi}{5}, 2\pi\}$.
Case $2$: $\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}$.
Case $3$: $\cos \frac{x}{2} = 0 \Rightarrow \frac{x}{2} = \frac{\pi}{2} \Rightarrow x = \pi$.
Sum of all values $= (0 + \frac{2\pi}{5} + \frac{4\pi}{5} + \frac{6\pi}{5} + \frac{8\pi}{5} + 2\pi) + (\frac{\pi}{2} + \frac{3\pi}{2}) + \pi = 6\pi + 2\pi + \pi = 9\pi$.

(D) Given equation: $\cos \left(\frac{\pi}{3}+x\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x$.
Using the identity $\cos(A+B)\cos(A-B) = \cos^2 A - \sin^2 B$,we get:
$\cos^2 \left(\frac{\pi}{3}\right) - \sin^2 x = \frac{1}{4} \cos^2 2x$.
Substituting $\cos^2 \left(\frac{\pi}{3}\right) = \frac{1}{4}$:
$\frac{1}{4} - \sin^2 x = \frac{1}{4} \cos^2 2x$.
Multiplying by $4$:
$1 - 4 \sin^2 x = \cos^2 2x$.
Using $1 - 2 \sin^2 x = \cos 2x$,we have $4 \sin^2 x = 2(1 - \cos 2x)$:
$1 - 2(1 - \cos 2x) = \cos^2 2x$.
$1 - 2 + 2 \cos 2x = \cos^2 2x$.
$\cos^2 2x - 2 \cos 2x + 1 = 0$.
$(\cos 2x - 1)^2 = 0$.
$\cos 2x = 1$.
$2x = 2n\pi \implies x = n\pi$.
For $x \in [-3\pi, 3\pi]$,the values of $n$ are $\{-3, -2, -1, 0, 1, 2, 3\}$.
There are $7$ such values.

(D) Given equation: $14 \operatorname{cosec}^{2} x - 2 \sin^{2} x = 21 - 4 \cos^{2} x$
Substitute $\cos^{2} x = 1 - \sin^{2} x$:
$14 \operatorname{cosec}^{2} x - 2 \sin^{2} x = 21 - 4(1 - \sin^{2} x)$
$14 \operatorname{cosec}^{2} x - 2 \sin^{2} x = 17 + 4 \sin^{2} x$
$14 \operatorname{cosec}^{2} x - 6 \sin^{2} x = 17$
Let $\sin^{2} x = p$,then $\operatorname{cosec}^{2} x = \frac{1}{p}$:
$\frac{14}{p} - 6p = 17 \implies 14 - 6p^{2} = 17p$
$6p^{2} + 17p - 14 = 0$
$(6p - 4)(p + 3.5) = 0$ (or using quadratic formula):
$p = \frac{2}{3}$ or $p = -3.5$ (rejected as $\sin^{2} x \geq 0$)
So,$\sin^{2} x = \frac{2}{3} \implies \sin x = \pm \sqrt{\frac{2}{3}}$
In the interval $\left(\frac{\pi}{4}, \frac{7 \pi}{4}\right)$,$\sin x$ takes the value $\sqrt{\frac{2}{3}}$ twice and $-\sqrt{\frac{2}{3}}$ twice.
Thus,there are $4$ solutions.

(B) Given the equation $\sin x = \cos^{2} x$.
Using the identity $\cos^{2} x = 1 - \sin^{2} x$,we get $\sin x = 1 - \sin^{2} x$.
Rearranging the terms,we obtain the quadratic equation $\sin^{2} x + \sin x - 1 = 0$.
Let $t = \sin x$. Then $t^{2} + t - 1 = 0$.
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$,we find $t = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.
Since $-1 \le \sin x \le 1$,we must have $\sin x = \frac{-1 + \sqrt{5}}{2} \approx 0.618$ (which is positive) or $\sin x = \frac{-1 - \sqrt{5}}{2} \approx -1.618$ (which is impossible).
Thus,$\sin x = \frac{\sqrt{5} - 1}{2}$.
In the interval $(0, 10)$,the value $10$ radians is approximately $10 / 3.14 \approx 3.18\pi$.
Since $\sin x = k$ (where $0 < k < 1$) has $2$ solutions in each interval of length $2\pi$,and the interval $(0, 10)$ covers slightly more than $3\pi$ (specifically $0$ to $3.18\pi$),we count the solutions:
In $(0, 2\pi)$,there are $2$ solutions.
In $(2\pi, 3\pi)$,there are $0$ solutions (as $\sin x$ is negative).
In $(3\pi, 3.18\pi)$,there are $2$ solutions.
Total number of solutions is $2 + 2 = 4$.

(A) The given equation is $2 \theta - \cos^{2} \theta + \sqrt{2} = 0$.
This can be rewritten as $\cos^{2} \theta = 2 \theta + \sqrt{2}$.
Let $f(\theta) = \cos^{2} \theta$ and $g(\theta) = 2 \theta + \sqrt{2}$.
The function $f(\theta) = \cos^{2} \theta$ is a periodic function with a range of $[0, 1]$.
The function $g(\theta) = 2 \theta + \sqrt{2}$ is a straight line with a slope of $2$ and a $y$-intercept of $\sqrt{2} \approx 1.414$.
Since the maximum value of $\cos^{2} \theta$ is $1$,and for $\theta > 0$,$g(\theta) > \sqrt{2} > 1$,there are no solutions for $\theta > 0$.
For $\theta < 0$,the line $g(\theta)$ intersects the curve $f(\theta)$ at exactly one point,as shown in the graph.

(C) We need to solve $|\cos x| = \sin x$ for $x \in [-4 \pi, 4 \pi]$.
Since $|\cos x| \geq 0$,we must have $\sin x \geq 0$. This implies $x$ must be in the first or second quadrant.
Case $1$: $\cos x = \sin x \implies \tan x = 1$. In $[0, 2 \pi]$,this gives $x = \frac{\pi}{4}, \frac{5 \pi}{4}$. Since $\sin x$ must be positive,only $x = \frac{\pi}{4}$ is a solution.
Case $2$: $-\cos x = \sin x \implies \tan x = -1$. In $[0, 2 \pi]$,this gives $x = \frac{3 \pi}{4}, \frac{7 \pi}{4}$. Since $\sin x$ must be positive,only $x = \frac{3 \pi}{4}$ is a solution.
Thus,there are $2$ solutions in every interval of length $2 \pi$ (specifically $[0, 2 \pi]$ and $[-2 \pi, 0]$).
In the interval $[-4 \pi, 4 \pi]$,which consists of $4$ intervals of length $2 \pi$,the total number of solutions is $2 \times 4 = 8$.

(A) Given equations are $2 \sin^{2} \theta - \cos 2\theta = 0$ and $2 \cos^{2} \theta + 3 \sin \theta = 0$.
For the first equation:
$2 \sin^{2} \theta - (1 - 2 \sin^{2} \theta) = 0$
$4 \sin^{2} \theta = 1$
$\sin^{2} \theta = \frac{1}{4} \implies \sin \theta = \pm \frac{1}{2}$.
In $[0, 2\pi]$,$\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$.
For the second equation:
$2(1 - \sin^{2} \theta) + 3 \sin \theta = 0$
$2 - 2 \sin^{2} \theta + 3 \sin \theta = 0$
$2 \sin^{2} \theta - 3 \sin \theta - 2 = 0$
$(2 \sin \theta + 1)(\sin \theta - 2) = 0$.
Since $\sin \theta = 2$ is impossible,we have $\sin \theta = -\frac{1}{2}$.
In $[0, 2\pi]$,$\theta = \frac{7\pi}{6}, \frac{11\pi}{6}$.
The common solutions are $\theta = \frac{7\pi}{6}, \frac{11\pi}{6}$.
Sum of solutions $= \frac{7\pi}{6} + \frac{11\pi}{6} = \frac{18\pi}{6} = 3\pi$.
Given sum $= k\pi$,so $k = 3$.

(B) Given equation: $\tan \theta(1 + \sqrt{5} \tan 2\theta) = \sqrt{5} - \tan 2\theta$
$\tan \theta + \sqrt{5} \tan \theta \tan 2\theta = \sqrt{5} - \tan 2\theta$
$\tan \theta + \tan 2\theta = \sqrt{5}(1 - \tan \theta \tan 2\theta)$
$\frac{\tan \theta + \tan 2\theta}{1 - \tan \theta \tan 2\theta} = \sqrt{5}$
$\tan(3\theta) = \sqrt{5}$
Let $\tan \alpha = \sqrt{5}$,where $\alpha \in (0, \pi/2)$. Then $3\theta = n\pi + \alpha$,so $\theta = \frac{n\pi}{3} + \frac{\alpha}{3}$.
Since $\theta \in [-\pi, \pi/2)$,we check values of $n$:
For $n = -3: \theta = -\pi + \alpha/3$ (Valid)
For $n = -2: \theta = -2\pi/3 + \alpha/3$ (Valid)
For $n = -1: \theta = -\pi/3 + \alpha/3$ (Valid)
For $n = 0: \theta = \alpha/3$ (Valid)
For $n = 1: \theta = \pi/3 + \alpha/3$ (Valid,since $\alpha/3 < \pi/6$,$\theta < \pi/2$)
We must exclude values where $\tan \theta$ or $\tan 2\theta$ are undefined or lead to division by zero. The set $S$ excludes $\pm \pi/2, \pm \pi/4, -3\pi/4$.
All $5$ values are within the domain $S$. Thus,the number of elements is $5$.

(B) Given equation: $\sin(9x) + \sin(3x) = 0$
Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$2 \sin(6x) \cos(3x) = 0$
This implies $\sin(6x) = 0$ or $\cos(3x) = 0$.
Case $1$: $\sin(6x) = 0$ $\Rightarrow 6x = n\pi$ $\Rightarrow x = \frac{n\pi}{6}$ for $n \in \mathbb{Z}$.
For $x \in [0, 2\pi]$,$n$ can be $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$.
Values: $x \in \{0, \frac{\pi}{6}, \frac{2\pi}{6}, \frac{3\pi}{6}, \frac{4\pi}{6}, \frac{5\pi}{6}, \frac{6\pi}{6}, \frac{7\pi}{6}, \frac{8\pi}{6}, \frac{9\pi}{6}, \frac{10\pi}{6}, \frac{11\pi}{6}, \frac{12\pi}{6}\}$ (Total $13$ values).
Case $2$: $\cos(3x) = 0$ $\Rightarrow 3x = (2k+1)\frac{\pi}{2}$ $\Rightarrow x = (2k+1)\frac{\pi}{6}$ for $k \in \mathbb{Z}$.
For $x \in [0, 2\pi]$,$k$ can be $0, 1, 2, 3, 4, 5$.
Values: $x \in \{\frac{\pi}{6}, \frac{3\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{9\pi}{6}, \frac{11\pi}{6}\}$.
All these values are already included in the set from Case $1$.
Thus,the total number of distinct solutions is $13$.

(C) Given equation: $\sin(x+x^2) - \sin(x^2) = \sin x$
Using the identity $\sin A - \sin B = 2 \sin(\frac{A-B}{2}) \cos(\frac{A+B}{2})$,we get:
$2 \sin(\frac{x}{2}) \cos(\frac{2x^2+x}{2}) = \sin x$
Since $\sin x = 2 \sin(\frac{x}{2}) \cos(\frac{x}{2})$,the equation becomes:
$2 \sin(\frac{x}{2}) \cos(\frac{2x^2+x}{2}) = 2 \sin(\frac{x}{2}) \cos(\frac{x}{2})$
This implies $2 \sin(\frac{x}{2}) [\cos(\frac{2x^2+x}{2}) - \cos(\frac{x}{2})] = 0$
Case $1$: $\sin(\frac{x}{2}) = 0$ $\Rightarrow \frac{x}{2} = n\pi$ $\Rightarrow x = 2n\pi$. For $n=1$,$x=2\pi \approx 6.28$,which is outside $[2, 3]$.
Case $2$: $\cos(\frac{2x^2+x}{2}) = \cos(\frac{x}{2}) \Rightarrow \frac{2x^2+x}{2} = 2k\pi \pm \frac{x}{2}$
Subcase $2a$: $x^2 + \frac{x}{2} = 2k\pi + \frac{x}{2} \Rightarrow x^2 = 2k\pi$. For $k=1$,$x = \sqrt{2\pi} \approx \sqrt{6.28} \approx 2.506$ (in $[2, 3]$).
Subcase $2b$: $x^2 + \frac{x}{2} = 2k\pi - \frac{x}{2} \Rightarrow x^2 + x - 2k\pi = 0$. For $k=1$,$x^2 + x - 2\pi = 0$. Roots are $x = \frac{-1 \pm \sqrt{1 + 8\pi}}{2}$. Taking positive root,$x = \frac{-1 + \sqrt{1 + 25.13}}{2} \approx \frac{-1 + 5.11}{2} \approx 2.055$ (in $[2, 3]$).
Thus,there are $2$ solutions.

(B) Given equation: $\sin \theta + \cos \theta = \sin 2\theta$.
Let $t = \sin \theta + \cos \theta$. Then $t^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1 + \sin 2\theta$.
So,$\sin 2\theta = t^2 - 1$.
The equation becomes $t = t^2 - 1$,or $t^2 - t - 1 = 0$.
Solving for $t$,we get $t = \frac{1 \pm \sqrt{5}}{2}$.
Since $t = \sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \frac{\pi}{4})$,the range of $t$ is $[-\sqrt{2}, \sqrt{2}]$.
Note that $\sqrt{2} \approx 1.414$ and $\frac{1 + \sqrt{5}}{2} \approx 1.618$. Since $1.618 > 1.414$,the value $t = \frac{1 + \sqrt{5}}{2}$ is rejected.
Thus,we must have $\sqrt{2} \sin(\theta + \frac{\pi}{4}) = \frac{1 - \sqrt{5}}{2}$.
$\sin(\theta + \frac{\pi}{4}) = \frac{1 - \sqrt{5}}{2\sqrt{2}} \approx \frac{-1.236}{2.828} \approx -0.437$.
Since $-1 < -0.437 < 0$,there are exactly two solutions for $\theta + \frac{\pi}{4}$ in any interval of length $2\pi$.
In the interval $[-\pi, \pi]$,the shifted interval for $\theta + \frac{\pi}{4}$ is $[-\frac{3\pi}{4}, \frac{5\pi}{4}]$,which has length $2\pi$.
Therefore,there are $2$ solutions.

(B) Given the equation $2 \sin 3x + \sin 7x - 3 = 0$.
This can be rewritten as $2 \sin 3x + \sin 7x = 3$.
Since the maximum value of $\sin \theta$ is $1$,the sum $2 \sin 3x + \sin 7x$ can be $3$ only if $\sin 3x = 1$ and $\sin 7x = 1$ simultaneously.
For $\sin 3x = 1$,$3x = 2n\pi + \frac{\pi}{2} \Rightarrow x = \frac{2n\pi}{3} + \frac{\pi}{6} = \frac{(4n+1)\pi}{6}$ for $n \in \mathbb{Z}$.
For $\sin 7x = 1$,$7x = 2m\pi + \frac{\pi}{2} \Rightarrow x = \frac{2m\pi}{7} + \frac{\pi}{14} = \frac{(4m+1)\pi}{14}$ for $m \in \mathbb{Z}$.
Equating the two expressions for $x$: $\frac{4n+1}{6} = \frac{4m+1}{14}$ $\Rightarrow 14(4n+1) = 6(4m+1)$ $\Rightarrow 56n + 14 = 24m + 6$ $\Rightarrow 56n + 8 = 24m$ $\Rightarrow 7n + 1 = 3m$.
Testing values for $n$ in the interval $[-2\pi, 2\pi]$ (i.e.,$x \in [-6.28, 6.28]$):
If $n = 0, x = \frac{\pi}{6} \approx 0.52$ (Valid,$3m = 1$ not integer).
If $n = 1, x = \frac{5\pi}{6} \approx 2.61$ (Valid,$3m = 8$ not integer).
If $n = 2, x = \frac{9\pi}{6} = 1.5\pi$ (Valid,$3m = 15 \Rightarrow m = 5$).
If $n = -2, x = -1.5\pi$ (Valid,$3m = -13$ not integer).
Checking common solutions: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{9\pi}{6}, \dots$ and $x = \frac{\pi}{14}, \frac{5\pi}{14}, \dots$. The common solutions in $[-2\pi, 2\pi]$ are $x = \frac{3\pi}{2}$ and $x = -\frac{3\pi}{2}$.

(B) Given equation: $8 \sin^3 \theta - 7 \sin \theta + \sqrt{3} \cos \theta = 0$
Using the identity $4 \sin^3 \theta = 3 \sin \theta - \sin 3 \theta$,we substitute:
$2(3 \sin \theta - \sin 3 \theta) - 7 \sin \theta + \sqrt{3} \cos \theta = 0$
$6 \sin \theta - 2 \sin 3 \theta - 7 \sin \theta + \sqrt{3} \cos \theta = 0$
$\sqrt{3} \cos \theta - \sin \theta = 2 \sin 3 \theta$
Dividing by $2$:
$\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta = \sin 3 \theta$
$\sin(60^{\circ} - \theta) = \sin 3 \theta$
$60^{\circ} - \theta = 3 \theta \implies 4 \theta = 60^{\circ} \implies \theta = 15^{\circ}$
Since $15^{\circ}$ lies in the interval $(10^{\circ}, 20^{\circ})$,the correct option is $B$.

(A) Given the equation $\cos(\sin x) = \sin(\cos x)$.
We know that $\sin(\theta) = \cos(\frac{\pi}{2} - \theta)$.
So,$\cos(\sin x) = \cos(\frac{\pi}{2} - \cos x)$.
This implies $\sin x = 2n\pi \pm (\frac{\pi}{2} - \cos x)$ for $n \in \mathbb{Z}$.
Case $1$: $\sin x + \cos x = 2n\pi + \frac{\pi}{2}$.
The range of $\sin x + \cos x$ is $[-\sqrt{2}, \sqrt{2}] \approx [-1.414, 1.414]$.
For $n=0$,$\frac{\pi}{2} \approx 1.57$,which is outside the range.
Case $2$: $\sin x - \cos x = 2n\pi - \frac{\pi}{2}$.
The range of $\sin x - \cos x$ is $[-\sqrt{2}, \sqrt{2}] \approx [-1.414, 1.414]$.
For $n=0$,$-\frac{\pi}{2} \approx -1.57$,which is outside the range.
Since no value of $n$ satisfies the equation,the number of elements in $X$ is $0$.

(B) Given equation: $\cos^4 x + \frac{1}{\cos^2 x} = \sin^4 x + \frac{1}{\sin^2 x}$
Rearranging the terms: $\cos^4 x - \sin^4 x = \frac{1}{\sin^2 x} - \frac{1}{\cos^2 x}$
Using the identity $a^2 - b^2 = (a-b)(a+b)$: $(\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x}$
Since $\cos^2 x + \sin^2 x = 1$,we have: $\cos^2 x - \sin^2 x = \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x}$
This implies $(\cos^2 x - \sin^2 x) \left(1 - \frac{1}{\sin^2 x \cos^2 x}\right) = 0$
Using $\sin 2x = 2 \sin x \cos x$,we get $\sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}$.
So,$\cos 2x \left(1 - \frac{4}{\sin^2 2x}\right) = 0$
This gives $\cos 2x = 0$ or $\sin^2 2x = 4$. Since $\sin^2 2x$ cannot be $4$,we must have $\cos 2x = 0$.
Thus,$2x = (2n+1)\frac{\pi}{2}$,which means $x = (2n+1)\frac{\pi}{4}$.
For $x \in [0, 2\pi]$,the solutions are $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
Therefore,the total number of solutions is $4$.

(C) Given equation: $\sin x + \frac{1}{2} \cos x = \sin^2(x + \frac{\pi}{4})$
Multiply by $2$: $2 \sin x + \cos x = 2 \sin^2(x + \frac{\pi}{4})$
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$2 \sin x + \cos x = 2 [\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}]^2$
$2 \sin x + \cos x = 2 [\frac{1}{\sqrt{2}}(\sin x + \cos x)]^2$
$2 \sin x + \cos x = 2 \times \frac{1}{2}(\sin x + \cos x)^2$
$2 \sin x + \cos x = \sin^2 x + \cos^2 x + 2 \sin x \cos x$
Since $\sin^2 x + \cos^2 x = 1$:
$2 \sin x + \cos x = 1 + 2 \sin x \cos x$
$2 \sin x - 2 \sin x \cos x + \cos x - 1 = 0$
$2 \sin x(1 - \cos x) - 1(1 - \cos x) = 0$
$(2 \sin x - 1)(1 - \cos x) = 0$
This gives $\sin x = \frac{1}{2}$ or $\cos x = 1$.
For $x \in [0, \pi]$,$\sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6}$.
For $x \in [0, \pi]$,$\cos x = 1 \Rightarrow x = 0$.
The sum of all such $x$ is $\frac{\pi}{6} + \frac{5\pi}{6} + 0 = \pi$.

(A) Given the equation: $\cos^7 \theta - \sin^4 \theta = 1$
Rearranging the terms,we get: $\cos^7 \theta = 1 + \sin^4 \theta$
We know that for any $\theta$,the range of $\cos^7 \theta$ is $[-1, 1]$.
Also,since $\sin^4 \theta \geq 0$,the right-hand side $(RHS)$ satisfies $1 + \sin^4 \theta \geq 1$.
For the equality to hold,both sides must be equal to $1$.
Thus,$\cos^7 \theta = 1$ and $\sin^4 \theta = 0$.
From $\cos^7 \theta = 1$,we get $\cos \theta = 1$,which implies $\theta = 0, 2\pi$ in the interval $[0, 2\pi]$.
From $\sin^4 \theta = 0$,we get $\sin \theta = 0$,which implies $\theta = 0, \pi, 2\pi$ in the interval $[0, 2\pi]$.
The common values of $\theta$ satisfying both conditions are $\theta = 0$ and $\theta = 2\pi$.
Therefore,there are $2$ roots in the given interval.

(A) Given equation: $\cos 2 \theta \cos \frac{\theta}{2} = \cos 3 \theta \cos \frac{9 \theta}{2}$.
Multiply by $2$ on both sides: $2 \cos 2 \theta \cos \frac{\theta}{2} = 2 \cos \frac{9 \theta}{2} \cos 3 \theta$.
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we get:
$\cos \frac{5 \theta}{2} + \cos \frac{3 \theta}{2} = \cos \frac{15 \theta}{2} + \cos \frac{3 \theta}{2}$.
$\cos \frac{15 \theta}{2} = \cos \frac{5 \theta}{2}$.
General solution: $\frac{15 \theta}{2} = 2 k \pi \pm \frac{5 \theta}{2}$.
Case $1$: $\frac{15 \theta}{2} - \frac{5 \theta}{2} = 2 k \pi$ $\Rightarrow 5 \theta = 2 k \pi$ $\Rightarrow \theta = \frac{2 k \pi}{5}$.
Case $2$: $\frac{15 \theta}{2} + \frac{5 \theta}{2} = 2 k \pi$ $\Rightarrow 10 \theta = 2 k \pi$ $\Rightarrow \theta = \frac{k \pi}{5}$.
Combining both,$\theta = \frac{k \pi}{5}$ for $k \in \mathbb{Z}$.
In $[-\pi, \pi]$,$\theta \in \{-\pi, -\frac{4 \pi}{5}, -\frac{3 \pi}{5}, -\frac{2 \pi}{5}, -\frac{\pi}{5}, 0, \frac{\pi}{5}, \frac{2 \pi}{5}, \frac{3 \pi}{5}, \frac{4 \pi}{5}, \pi\}$.
Positive values $(m)$: $\{\frac{\pi}{5}, \frac{2 \pi}{5}, \frac{3 \pi}{5}, \frac{4 \pi}{5}, \pi\}$,so $m = 5$.
Negative values $(n)$: $\{-\pi, -\frac{4 \pi}{5}, -\frac{3 \pi}{5}, -\frac{2 \pi}{5}, -\frac{\pi}{5}\}$,so $n = 5$.
Therefore,$mn = 5 \times 5 = 25$.

(D) Given equation: $\log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1$
Let $u = \ln \sin x$ and $v = \ln \cos x$. Then $\cot x = \frac{e^v}{e^u}$ and $\tan x = \frac{e^u}{e^v}$.
The equation becomes: $\frac{v-u}{v} + 4\frac{u-v}{u} = 1$
$1 - \frac{u}{v} + 4\frac{u}{v} - 4 = 1$
$3\frac{u}{v} = 4$ $\Rightarrow 3u = 4v$ $\Rightarrow 3 \ln \sin x = 4 \ln \cos x$
$\sin^3 x = \cos^4 x = (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x$
$\sin^4 x + \sin^3 x - 2\sin^2 x - 1 = 0$. This approach is complex. Let's re-evaluate the original equation: $\frac{\ln \cos x - \ln \sin x}{\ln \cos x} + 4 \frac{\ln \sin x - \ln \cos x}{\ln \sin x} = 1$
Let $a = \ln \sin x$ and $b = \ln \cos x$. Then $\frac{b-a}{b} + 4\frac{a-b}{a} = 1$
$1 - \frac{a}{b} + 4 - 4\frac{b}{a} = 1 \Rightarrow 4 = \frac{a}{b} + 4\frac{b}{a}$. Let $t = \frac{a}{b}$.
$t + \frac{4}{t} = 4$ $\Rightarrow t^2 - 4t + 4 = 0$ $\Rightarrow (t-2)^2 = 0$ $\Rightarrow t = 2$.
So,$\frac{\ln \sin x}{\ln \cos x} = 2$ $\Rightarrow \ln \sin x = 2 \ln \cos x$ $\Rightarrow \sin x = \cos^2 x = 1 - \sin^2 x$.
$\sin^2 x + \sin x - 1 = 0$. Using the quadratic formula,$\sin x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 + \sqrt{5}}{2}$ (since $\sin x > 0$ for $x \in (0, \pi/2)$).
Comparing $\sin x = \frac{-1 + \sqrt{5}}{2}$ with $\sin x = \frac{\alpha + \sqrt{\beta}}{2}$,we get $\alpha = -1$ and $\beta = 5$.
Therefore,$\alpha + \beta = -1 + 5 = 4$.

(NONE) Given equation: $3 \cos^4 \theta - 5 \cos^2 \theta - 2 \sin^2 \theta + 2 = 0$.
Using $\sin^2 \theta = 1 - \cos^2 \theta$,we substitute:
$3 \cos^4 \theta - 5 \cos^2 \theta - 2(1 - \cos^2 \theta) + 2 = 0$
$3 \cos^4 \theta - 5 \cos^2 \theta - 2 + 2 \cos^2 \theta + 2 = 0$
$3 \cos^4 \theta - 3 \cos^2 \theta = 0$
$3 \cos^2 \theta (\cos^2 \theta - 1) = 0$
$3 \cos^2 \theta (-\sin^2 \theta) = 0$
$-3 \cos^2 \theta \sin^2 \theta = 0$
This implies $\cos^2 \theta = 0$ or $\sin^2 \theta = 0$.
Case $1$: $\cos^2 \theta = 0 \implies \cos \theta = 0$. In $[0, 2\pi]$,$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$.
Case $2$: $\sin^2 \theta = 0 \implies \sin \theta = 0$. In $[0, 2\pi]$,$\theta = 0, \pi, 2\pi$.
The set $S = \{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\}$.
The number of elements is $5$.

(C) Given equation: $4 \cos \theta + 5 \sin \theta = 1$.
Divide by $\cos \theta$ (since $\cos \theta \neq 0$): $4 + 5 \tan \theta = \sec \theta$.
Squaring both sides: $(4 + 5 \tan \theta)^2 = \sec^2 \theta = 1 + \tan^2 \theta$.
$16 + 25 \tan^2 \theta + 40 \tan \theta = 1 + \tan^2 \theta$.
$24 \tan^2 \theta + 40 \tan \theta + 15 = 0$.
Using the quadratic formula for $\tan \theta$: $\tan \theta = \frac{-40 \pm \sqrt{1600 - 4(24)(15)}}{2(24)} = \frac{-40 \pm \sqrt{1600 - 1440}}{48} = \frac{-40 \pm \sqrt{160}}{48} = \frac{-40 \pm 4\sqrt{10}}{48} = \frac{-10 \pm \sqrt{10}}{12}$.
Since $-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$,$\cos \alpha > 0$. From $4 \cos \alpha + 5 \sin \alpha = 1$,we have $5 \sin \alpha = 1 - 4 \cos \alpha$.
For $\alpha$ to be a solution,$\tan \alpha = \frac{-10 + \sqrt{10}}{12}$ satisfies the original equation. Thus,the correct option is $C$.