Solution of trigonometrical equations Questions in English

(A) Given equation: $(\sqrt{3}-1) \sin \theta + (\sqrt{3}+1) \cos \theta = 2$.
Divide both sides by $\sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} = \sqrt{(3-2\sqrt{3}+1) + (3+2\sqrt{3}+1)} = \sqrt{8} = 2\sqrt{2}$.
$\frac{\sqrt{3}-1}{2\sqrt{2}} \sin \theta + \frac{\sqrt{3}+1}{2\sqrt{2}} \cos \theta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Let $\cos \alpha = \frac{\sqrt{3}-1}{2\sqrt{2}}$ and $\sin \alpha = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
Then $\tan \alpha = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \tan(75^\circ) = \tan(\frac{5\pi}{12})$.
So,$\sin(\theta + \alpha) = \sin(\frac{\pi}{4})$.
Thus,$\theta + \frac{5\pi}{12} = n\pi + (-1)^n \frac{\pi}{4}$.
$\theta = n\pi + (-1)^n \frac{\pi}{4} - \frac{5\pi}{12}$.
Alternatively,using $\cos(\theta - \beta) = \frac{1}{\sqrt{2}}$ where $\beta = \frac{\pi}{12}$,we get $\theta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$.

(A) Given the equation: $\cot \frac{x}{2} - \operatorname{cosec} \frac{x}{2} = \cot x$.
Using the identities $\cot \theta = \frac{\cos \theta}{\sin \theta}$ and $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$,we have:
$\frac{\cos(x/2) - 1}{\sin(x/2)} = \frac{\cos x}{\sin x}$.
Using $\cos(x/2) - 1 = -2 \sin^2(x/4)$ and $\sin(x/2) = 2 \sin(x/4) \cos(x/4)$,the left side becomes:
$\frac{-2 \sin^2(x/4)}{2 \sin(x/4) \cos(x/4)} = -\tan(x/4)$.
The right side is $\cot x = \frac{1}{\tan x}$.
So,$-\tan(x/4) = \frac{1}{\tan x}$,which implies $\tan x \cdot \tan(x/4) = -1$.
This equation has no real solutions for $x$ because $\tan x$ and $\tan(x/4)$ cannot satisfy this product condition for all $n \in \mathbb{Z}$. However,checking the options provided,none satisfy the equation. Given the standard form of such problems,there is no solution.

(B) Given trigonometric equations are $\tan \theta = -1$ and $\cos \theta = \frac{1}{\sqrt{2}}$.
Since $\tan \theta$ is negative and $\cos \theta$ is positive,$\theta$ must lie in the fourth quadrant.
The general solution for $\tan \theta = -1$ is $\theta = n \pi + \frac{3 \pi}{4}$.
The general solution for $\cos \theta = \frac{1}{\sqrt{2}}$ is $\theta = 2 n \pi \pm \frac{\pi}{4}$.
For $\theta$ to satisfy both,we look for the angle in the fourth quadrant where $\tan \theta = -1$ and $\cos \theta = \frac{1}{\sqrt{2}}$,which is $\theta = \frac{7 \pi}{4}$.
Thus,the general solution is $\theta = 2 n \pi + \frac{7 \pi}{4}$.

(B) Given equation: $2 \sin^2 \theta - \sin \theta \cos \theta - 3 \cos^2 \theta = 0$.
Dividing by $\cos^2 \theta$ (assuming $\cos \theta \neq 0$):
$2 \tan^2 \theta - \tan \theta - 3 = 0$.
Let $x = \tan \theta$,then $2x^2 - x - 3 = 0$.
$(2x - 3)(x + 1) = 0$.
So,$\tan \theta = \frac{3}{2}$ or $\tan \theta = -1$.
For $\tan \theta = -1$,$\theta = -\frac{\pi}{4}$ and $\theta = \frac{3\pi}{4}$ in $(-\pi, \pi)$.
For $\tan \theta = \frac{3}{2}$,$\theta = \arctan(\frac{3}{2})$ and $\theta = \arctan(\frac{3}{2}) - \pi$ in $(-\pi, \pi)$.
Total number of solutions is $4$.

(D) Given equation: $\cos \theta + \cos 2\theta - \sqrt{3}(\sin \theta + \sin 2\theta) + 1 = 0$.
Using $\cos 2\theta = 2\cos^2 \theta - 1$ and $\sin 2\theta = 2\sin \theta \cos \theta$:
$\cos \theta + 2\cos^2 \theta - 1 - \sqrt{3}\sin \theta - 2\sqrt{3}\sin \theta \cos \theta + 1 = 0$.
$\cos \theta(1 + 2\cos \theta) - \sqrt{3}\sin \theta(1 + 2\cos \theta) = 0$.
$(1 + 2\cos \theta)(\cos \theta - \sqrt{3}\sin \theta) = 0$.
Case $1$: $1 + 2\cos \theta = 0 \implies \cos \theta = -1/2$.
In $(0, 2\pi)$,$\theta = 2\pi/3, 4\pi/3$.
Case $2$: $\cos \theta - \sqrt{3}\sin \theta = 0 \implies \tan \theta = 1/\sqrt{3}$.
In $(0, 2\pi)$,$\theta = \pi/6, 7\pi/6$.
The solutions are $\pi/6, 2\pi/3, 7\pi/6, 4\pi/3$.
Total number of solutions is $4$.

(B) Given the equation $\sec x \cos 5x + 1 = 0$,we can rewrite it as $\frac{\cos 5x}{\cos x} = -1$,provided $\cos x \neq 0$.
This implies $\cos 5x = -\cos x$,which is equivalent to $\cos 5x = \cos(\pi - x)$.
The general solution for $\cos \theta = \cos \alpha$ is $\theta = 2n\pi \pm \alpha$.
Case $1$: $5x = 2n\pi + (\pi - x) \implies 6x = (2n+1)\pi \implies x = \frac{(2n+1)\pi}{6}$.
For $x \in [0, 2\pi]$,$n = 0, 1, 2, 3, 4, 5$ gives $x = \frac{\pi}{6}, \frac{3\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{9\pi}{6}, \frac{11\pi}{6}$.
Checking $\cos x \neq 0$: $\frac{3\pi}{6} = \frac{\pi}{2}$ and $\frac{9\pi}{6} = \frac{3\pi}{2}$ make $\cos x = 0$,so these are excluded.
Remaining values: $\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$ ($4$ solutions).
Case $2$: $5x = 2n\pi - (\pi - x) \implies 4x = (2n-1)\pi \implies x = \frac{(2n-1)\pi}{4}$.
For $x \in [0, 2\pi]$,$n = 1, 2, 3, 4$ gives $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
None of these make $\cos x = 0$ ($4$ solutions).
Total solutions = $4 + 4 = 8$.

(D) Given the equation $\cos x \sqrt{16 \sin ^2 x} = 1$.
This simplifies to $\cos x \cdot 4 |\sin x| = 1$,or $2 \sin(2x) = 1$ (when $\sin x > 0$) and $-2 \sin(2x) = 1$ (when $\sin x < 0$).
Case $1$: $\sin x > 0$ $(x \in (0, \pi))$. Then $2 \sin(2x) = 1 \implies \sin(2x) = \frac{1}{2}$.
$2x = \frac{\pi}{6}, \frac{5 \pi}{6} \implies x = \frac{\pi}{12}, \frac{5 \pi}{12}$. Both are in $(0, \pi)$.
Case $2$: $\sin x < 0$ $(x \in (\pi, 2 \pi))$. Then $-2 \sin(2x) = 1 \implies \sin(2x) = -\frac{1}{2}$.
$2x = \frac{7 \pi}{6}, \frac{11 \pi}{6} \implies x = \frac{7 \pi}{12}, \frac{11 \pi}{12}$. However,these values are in $(0, \pi)$,where $\sin x > 0$,so they are rejected.
Wait,checking $2x$ range for $x \in (\pi, 2 \pi)$ is $2x \in (2 \pi, 4 \pi)$.
$2x = 2 \pi + \frac{7 \pi}{6} = \frac{19 \pi}{6} \implies x = \frac{19 \pi}{12}$ and $2x = 2 \pi + \frac{11 \pi}{6} = \frac{23 \pi}{6} \implies x = \frac{23 \pi}{12}$.
Sum of solutions $= \frac{\pi}{12} + \frac{5 \pi}{12} + \frac{19 \pi}{12} + \frac{23 \pi}{12} = \frac{48 \pi}{12} = 4 \pi$.

(A) Given equation: $4 \cos 2 \theta \cos 3 \theta = \sec \theta$
Multiply by $\cos \theta$ (assuming $\cos \theta \neq 0$):
$4 \cos 2 \theta \cos 3 \theta \cos \theta = 1$
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$2 \cos 2 \theta (\cos 4 \theta + \cos 2 \theta) = 1$
$2 \cos 2 \theta \cos 4 \theta + 2 \cos^2 2 \theta = 1$
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ and $2 \cos^2 A = 1 + \cos 2A$:
$(\cos 6 \theta + \cos 2 \theta) + (1 + \cos 4 \theta) = 1$
$\cos 6 \theta + \cos 4 \theta + \cos 2 \theta = 0$
Using $\cos 6 \theta + \cos 2 \theta = 2 \cos 4 \theta \cos 2 \theta$:
$\cos 4 \theta (2 \cos 2 \theta + 1) = 0$
Case $1$: $\cos 4 \theta = 0 \implies 4 \theta = (2n+1) \frac{\pi}{2} \implies \theta = \frac{(2n+1) \pi}{8}$ for $n = 0, 1, \dots, 7$ ($8$ solutions).
Case $2$: $\cos 2 \theta = -\frac{1}{2} \implies 2 \theta = 2n \pi \pm \frac{2 \pi}{3} \implies \theta = n \pi \pm \frac{\pi}{3}$ for $n = 0, 1, 2$ ($4$ solutions).
Total solutions = $8 + 4 = 12$.

(B) Given equation: $\sin^2 \theta + 3 \cos^2 \theta = 5 \sin \theta$
Substitute $\cos^2 \theta = 1 - \sin^2 \theta$:
$\sin^2 \theta + 3(1 - \sin^2 \theta) = 5 \sin \theta$
$\sin^2 \theta + 3 - 3 \sin^2 \theta = 5 \sin \theta$
$-2 \sin^2 \theta - 5 \sin \theta + 3 = 0$
$2 \sin^2 \theta + 5 \sin \theta - 3 = 0$
Factor the quadratic equation:
$(2 \sin \theta - 1)(\sin \theta + 3) = 0$
This gives $\sin \theta = \frac{1}{2}$ or $\sin \theta = -3$.
Since $-1 \le \sin \theta \le 1$,we discard $\sin \theta = -3$.
For $\sin \theta = \frac{1}{2} = \sin \frac{\pi}{6}$,the general solution is $\theta = n \pi + (-1)^n \frac{\pi}{6}, n \in Z$.

(C) Given equation: $2 \tan 2 \theta - \cot 2 \theta + 1 = 0$.
Let $x = \tan 2 \theta$. Then $\cot 2 \theta = \frac{1}{x}$.
The equation becomes $2x - \frac{1}{x} + 1 = 0$,which simplifies to $2x^2 + x - 1 = 0$.
Factoring the quadratic: $(2x - 1)(x + 1) = 0$,so $x = \frac{1}{2}$ or $x = -1$.
Case $1$: $\tan 2 \theta = -1$.
$2 \theta = n \pi - \frac{\pi}{4} \Rightarrow \theta = \frac{n \pi}{2} - \frac{\pi}{8}$.
For $\theta \in [0, \pi]$,possible values are $\theta = \frac{3 \pi}{8}$ $(n=1)$ and $\theta = \frac{7 \pi}{8}$ $(n=2)$.
Case $2$: $\tan 2 \theta = \frac{1}{2}$.
$2 \theta = n \pi + \tan^{-1}(\frac{1}{2}) \Rightarrow \theta = \frac{n \pi}{2} + \frac{1}{2} \tan^{-1}(\frac{1}{2})$.
For $\theta \in [0, \pi]$,possible values are $\theta = \frac{1}{2} \tan^{-1}(\frac{1}{2})$ $(n=0)$ and $\theta = \frac{\pi}{2} + \frac{1}{2} \tan^{-1}(\frac{1}{2})$ $(n=1)$.
Thus,there are $2 + 2 = 4$ solutions in the interval $[0, \pi]$.

(A) Given equation: $\tan x + \tan 2x - \tan 3x = 0$
We know that $\tan 3x = \tan(x + 2x) = \frac{\tan x + \tan 2x}{1 - \tan x \tan 2x}$
So,$\tan x + \tan 2x = \tan 3x(1 - \tan x \tan 2x)$
$\tan x + \tan 2x = \tan 3x - \tan x \tan 2x \tan 3x$
$\tan x + \tan 2x - \tan 3x = -\tan x \tan 2x \tan 3x$
Since the given equation is $\tan x + \tan 2x - \tan 3x = 0$,we have:
$-\tan x \tan 2x \tan 3x = 0$
This implies $\tan x = 0$ or $\tan 2x = 0$ or $\tan 3x = 0$
For $\tan x = 0$,$x = n\pi$
For $\tan 2x = 0$,$2x = n\pi \Rightarrow x = \frac{n\pi}{2}$
For $\tan 3x = 0$,$3x = n\pi \Rightarrow x = \frac{n\pi}{3}$
Combining these,the set of solutions is $\left\{x \mid x = \frac{n\pi}{3}, n \in Z\right\}$.

(A) Given equation: $\sin \left(x+\frac{\pi}{3}\right)+\sin \left(x-\frac{\pi}{3}\right)=1$
Using the identity $\sin(A+B)+\sin(A-B) = 2 \sin A \cos B$:
$2 \sin x \cos \frac{\pi}{3} = 1$
Since $\cos \frac{\pi}{3} = \frac{1}{2}$,we have:
$2 \sin x \cdot \frac{1}{2} = 1$
$\sin x = 1$
For $x \in [0, \pi]$,the value of $x$ is $\frac{\pi}{2}$.

(D) Given $\sin \left(\frac{\pi}{4} \cot \theta\right)=\cos \left(\frac{\pi}{4} \tan \theta\right)$.
Using the identity $\cos x = \sin \left(\frac{\pi}{2} - x\right)$,
$\sin \left(\frac{\pi}{4} \cot \theta\right) = \sin \left(\frac{\pi}{2} - \frac{\pi}{4} \tan \theta\right)$.
Equating the arguments (general solution $\sin A = \sin B \implies A = n\pi + (-1)^n B$ is complex here,but we check the principal branch),
$\frac{\pi}{4} \cot \theta = \frac{\pi}{2} - \frac{\pi}{4} \tan \theta$.
$\frac{\pi}{4} (\cot \theta + \tan \theta) = \frac{\pi}{2}$.
$\cot \theta + \tan \theta = 2$.
$\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = 2$.
$\frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = 2$.
$\frac{1}{\sin \theta \cos \theta} = 2$.
$1 = 2 \sin \theta \cos \theta$.
$1 = \sin 2 \theta$.
Since $\sin 2 \theta = 1$,we have $2 \theta = 2n \pi + \frac{\pi}{2}$.
Dividing by $2$,we get $\theta = n \pi + \frac{\pi}{4}$.

(C) Given the equation $(\sin x + \cos x)^{1 + \sin 2x} = 2$.
Since $1 + \sin 2x = (\sin x + \cos x)^2$,the equation becomes $(\sin x + \cos x)^{(\sin x + \cos x)^2} = 2$.
Let $u = \sin x + \cos x$. Then $u^{u^2} = 2$.
We know that $-\sqrt{2} \leq \sin x + \cos x \leq \sqrt{2}$,so $-\sqrt{2} \leq u \leq \sqrt{2}$.
If $u = \sqrt{2}$,then $(\sqrt{2})^{(\sqrt{2})^2} = (\sqrt{2})^2 = 2$. This is a solution.
For $u = \sqrt{2}$,$\sin x + \cos x = \sqrt{2} \implies \frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x = 1 \implies \sin(x + \frac{\pi}{4}) = 1$.
Thus,$x + \frac{\pi}{4} = \frac{\pi}{2} \implies x = \frac{\pi}{4}$.
If $u = -\sqrt{2}$,then $(-\sqrt{2})^{(-\sqrt{2})^2} = (-\sqrt{2})^2 = 2$. This is also a solution.
For $u = -\sqrt{2}$,$\sin x + \cos x = -\sqrt{2} \implies \sin(x + \frac{\pi}{4}) = -1$.
Thus,$x + \frac{\pi}{4} = -\frac{\pi}{2} \implies x = -\frac{3\pi}{4}$.
Comparing with the given options,$x = \frac{\pi}{4}$ is the correct choice.

(C) Given the equation: $4 \sin^2(x) - 4 \sin(x) + 1 = 0$
This is a quadratic equation in terms of $\sin(x)$,which can be written as $(2 \sin(x) - 1)^2 = 0$.
Taking the square root on both sides,we get: $2 \sin(x) - 1 = 0$
$\sin(x) = \frac{1}{2}$
We know that $\sin(x) = \sin(\frac{\pi}{6})$.
The general solution for $\sin(x) = \sin(\alpha)$ is $x = n\pi + (-1)^n \alpha$,where $n \in \mathbb{Z}$.
Therefore,$x = n\pi + (-1)^n \frac{\pi}{6}, n \in \mathbb{Z}$.
Hence,option $C$ is correct.

(B) Given the equation: $\tan(x) + \sec(x) = \sqrt{3}$.
We know the identity $\sec^2(x) - \tan^2(x) = 1$,which implies $(\sec(x) - \tan(x))(\sec(x) + \tan(x)) = 1$.
Substituting the given value: $(\sec(x) - \tan(x))(\sqrt{3}) = 1$,so $\sec(x) - \tan(x) = \frac{1}{\sqrt{3}}$.
Adding the two equations:
$(\tan(x) + \sec(x)) + (\sec(x) - \tan(x)) = \sqrt{3} + \frac{1}{\sqrt{3}}$
$2\sec(x) = \frac{3+1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$ $\Rightarrow \sec(x) = \frac{2}{\sqrt{3}}$ $\Rightarrow \cos(x) = \frac{\sqrt{3}}{2}$.
For $x \in [0, 2\pi]$,the solutions are $x = \frac{\pi}{6}$ and $x = \frac{11\pi}{6}$.
Checking the original equation: $\tan(\frac{\pi}{6}) + \sec(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.
Thus,$x = \frac{\pi}{6}$ is a valid solution.

(B) Given equation: $|\tan \theta| = \tan \theta + \sec \theta$.
Case $1$: If $\tan \theta \ge 0$,then $\tan \theta = \tan \theta + \sec \theta$,which implies $\sec \theta = 0$. This has no solution.
Case $2$: If $\tan \theta < 0$,then $-\tan \theta = \tan \theta + \sec \theta$,which implies $2\tan \theta + \sec \theta = 0$.
Substituting $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$,we get $2\sin \theta + 1 = 0$,so $\sin \theta = -\frac{1}{2}$.
Since $\tan \theta < 0$,$\theta$ must be in the second or fourth quadrant. For $\sin \theta = -\frac{1}{2}$,$\theta = \frac{7\pi}{6}$ or $\theta = \frac{11\pi}{6}$.
Checking $\tan \theta < 0$:
For $\theta = \frac{7\pi}{6}$,$\tan \theta = \tan(\frac{7\pi}{6}) = \frac{1}{\sqrt{3}} > 0$ (Reject).
For $\theta = \frac{11\pi}{6}$,$\tan \theta = \tan(\frac{11\pi}{6}) = -\frac{1}{\sqrt{3}} < 0$ (Accept).
Thus,the solution is $\theta = \frac{11\pi}{6}$.

(A) Given the equation: $4(\sin 2x \sin 4x + \sin^2 x) = 3$
Using the identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ and $2 \sin^2 x = 1 - \cos 2x$:
$2(2 \sin 2x \sin 4x) + 2(2 \sin^2 x) = 3$
$2(\cos(4x-2x) - \cos(4x+2x)) + 2(1 - \cos 2x) = 3$
$2(\cos 2x - \cos 6x) + 2 - 2 \cos 2x = 3$
$2 \cos 2x - 2 \cos 6x + 2 - 2 \cos 2x = 3$
$-2 \cos 6x + 2 = 3$
$-2 \cos 6x = 1$
$\cos 6x = -\frac{1}{2}$
Since $\cos \theta = \cos \alpha \Rightarrow \theta = 2n\pi \pm \alpha$,where $\alpha = \frac{2\pi}{3}$:
$6x = 2n\pi \pm \frac{2\pi}{3}$
Dividing by $6$:
$x = \frac{n\pi}{3} \pm \frac{\pi}{9}, n \in Z$

(C) Given the equation: $|\cos x|^{2\sin^2 x - 3\sin x + 1} = 1$.
This equation holds if:
$1)$ $|\cos x| = 1$
$2)$ The exponent is $0$ (provided the base is not $0$).
Case $1$: $|\cos x| = 1 \implies \cos x = 1$ or $\cos x = -1$.
For $x \in [0, 2\pi]$,this gives $x = 0, \pi, 2\pi$.
Case $2$: $2\sin^2 x - 3\sin x + 1 = 0$.
$(2\sin x - 1)(\sin x - 1) = 0$.
$\sin x = \frac{1}{2}$ or $\sin x = 1$.
If $\sin x = \frac{1}{2}$,then $x = \frac{\pi}{6}, \frac{5\pi}{6}$.
If $\sin x = 1$,then $x = \frac{\pi}{2}$.
However,the domain is $x \in [0, 2\pi] - \{\frac{\pi}{2}, \frac{3\pi}{2}\}$.
Thus,$x = \frac{\pi}{2}$ is excluded.
Also,we must check if the base $|\cos x| = 0$ when the exponent is $0$.
If $x = \frac{\pi}{2}$ or $\frac{3\pi}{2}$,$|\cos x| = 0$,but these are excluded from the domain.
Combining the valid solutions: $x \in \{0, \pi, 2\pi, \frac{\pi}{6}, \frac{5\pi}{6}\}$.
The total number of values is $5$.

(C) Given equation: $\sqrt{3-5 \sin x+\sin ^2 x} = -\cos x$.
For the square root to be defined and equal to $-\cos x$,we must have $-\cos x \ge 0$,which implies $\cos x \le 0$.
Squaring both sides: $3-5 \sin x+\sin ^2 x = \cos ^2 x$.
Using $\cos ^2 x = 1 - \sin ^2 x$: $3-5 \sin x+\sin ^2 x = 1 - \sin ^2 x$.
$2 \sin ^2 x - 5 \sin x + 2 = 0$.
Factoring the quadratic: $(2 \sin x - 1)(\sin x - 2) = 0$.
This gives $\sin x = \frac{1}{2}$ or $\sin x = 2$.
Since $\sin x \in [-1, 1]$,we must have $\sin x = \frac{1}{2}$.
For $\sin x = \frac{1}{2}$,$x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$.
Since $\cos x \le 0$,we must have $x = \frac{5\pi}{6}$ in the interval $[0, 2\pi]$.
The general solution for $\sin x = \frac{1}{2}$ and $\cos x < 0$ is $x = 2n\pi + \frac{5\pi}{6}$,which is equivalent to $x = (2n+1)\pi - \frac{\pi}{6}$.

(B) Given the equation: $\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x$.
Group the terms: $(\sin 3x + \sin x) + \sin 2x = (\cos 3x + \cos x) + \cos 2x$.
Using sum-to-product formulas: $2 \sin 2x \cos x + \sin 2x = 2 \cos 2x \cos x + \cos 2x$.
Factor out common terms: $\sin 2x (2 \cos x + 1) = \cos 2x (2 \cos x + 1)$.
Rearrange: $(2 \cos x + 1)(\sin 2x - \cos 2x) = 0$.
Case $1$: $2 \cos x + 1 = 0 \implies \cos x = -1/2$. In the interval $[0, 2\pi]$,$x = 2\pi/3, 4\pi/3$.
Case $2$: $\sin 2x - \cos 2x = 0 \implies \tan 2x = 1$.
For $0 \leq x \leq 2\pi$,$0 \leq 2x \leq 4\pi$.
$2x = \pi/4, 5\pi/4, 9\pi/4, 13\pi/4$.
$x = \pi/8, 5\pi/8, 9\pi/8, 13\pi/8$.
Total values of $x$ are $2 + 4 = 6$.

(C) Given equation: $4 \cos 2 \theta \cos 3 \theta = \sec \theta$
Since $\sec \theta = \frac{1}{\cos \theta}$,we have $4 \cos 2 \theta \cos 3 \theta \cos \theta = 1$.
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we get $2 \cos 2 \theta (\cos 4 \theta + \cos 2 \theta) = 1$.
$2 \cos 2 \theta \cos 4 \theta + 2 \cos^2 2 \theta = 1$.
Using $2 \cos^2 2 \theta - 1 = \cos 4 \theta$,we get $2 \cos 2 \theta \cos 4 \theta + \cos 4 \theta = 0$.
$\cos 4 \theta (2 \cos 2 \theta + 1) = 0$.
Case $1$: $\cos 4 \theta = 0 \Rightarrow 4 \theta = (2n + 1) \frac{\pi}{2} \Rightarrow \theta = (2n + 1) \frac{\pi}{8}$.
For $0 < \theta < \pi$,$\theta \in \{ \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \}$.
Case $2$: $2 \cos 2 \theta + 1 = 0 \Rightarrow \cos 2 \theta = -\frac{1}{2}$.
$2 \theta = 2n\pi \pm \frac{2\pi}{3} \Rightarrow \theta = n\pi \pm \frac{\pi}{3}$.
For $0 < \theta < \pi$,$\theta \in \{ \frac{\pi}{3}, \frac{2\pi}{3} \}$.
Checking $\cos \theta \neq 0$: None of these values make $\cos \theta = 0$.
Total solutions = $4 + 2 = 6$.

(C) Given the equation: $1 + \cos^2 \theta = 3 \sin \theta \cos \theta$.
Divide both sides by $\cos^2 \theta$ (assuming $\cos \theta \neq 0$):
$\sec^2 \theta + 1 = 3 \tan \theta$.
Since $\sec^2 \theta = 1 + \tan^2 \theta$,we have:
$(1 + \tan^2 \theta) + 1 = 3 \tan \theta$
$\tan^2 \theta - 3 \tan \theta + 2 = 0$.
Factoring the quadratic:
$(\tan \theta - 1)(\tan \theta - 2) = 0$.
This gives two cases:
Case $1$: $\tan \theta = 1 \Rightarrow \theta = n\pi + \frac{\pi}{4}$.
Case $2$: $\tan \theta = 2 \Rightarrow \theta = n\pi + \tan^{-1}(2)$.
Thus,the solution is $n\pi + \frac{\pi}{4}, n\pi + \tan^{-1}(2); n \in \mathbb{Z}$.

(A) Given equation: $\sin x + \sin 2x + \sin 3x + \sin 4x = 0$.
Grouping terms: $(\sin 4x + \sin x) + (\sin 3x + \sin 2x) = 0$.
Using the formula $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$2 \sin(\frac{5x}{2}) \cos(\frac{3x}{2}) + 2 \sin(\frac{5x}{2}) \cos(\frac{x}{2}) = 0$.
$2 \sin(\frac{5x}{2}) [\cos(\frac{3x}{2}) + \cos(\frac{x}{2})] = 0$.
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 $1$: $\sin(\frac{5x}{2}) = 0 \implies \frac{5x}{2} = n \pi \implies x = \frac{2n \pi}{5}$. For $0 \leq x \leq 2 \pi$,$x \in \{0, \frac{2 \pi}{5}, \frac{4 \pi}{5}, \frac{6 \pi}{5}, \frac{8 \pi}{5}, 2 \pi\}$.
Case $2$: $\cos x = 0 \implies x = \frac{\pi}{2}, \frac{3 \pi}{2}$.
Case $3$: $\cos(\frac{x}{2}) = 0 \implies \frac{x}{2} = \frac{\pi}{2} \implies x = \pi$.
Combining all unique values: $\{0, \frac{2 \pi}{5}, \frac{\pi}{2}, \frac{4 \pi}{5}, \pi, \frac{6 \pi}{5}, \frac{3 \pi}{2}, \frac{8 \pi}{5}, 2 \pi\}$.
Total number of values is $9$.

(C) Given equation: $1+\cos x \cdot \cos 5 x=\sin ^2 x$
Using the identity $\sin ^2 x = 1 - \cos ^2 x$,we get:
$1+\cos x \cdot \cos 5 x = 1 - \cos ^2 x$
$\Rightarrow \cos ^2 x + \cos x \cdot \cos 5 x = 0$
$\Rightarrow \cos x(\cos x + \cos 5 x) = 0$
Using the sum-to-product formula $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$\Rightarrow \cos x [2 \cos(3x) \cos(-2x)] = 0$
Since $\cos(-2x) = \cos(2x)$,we have:
$2 \cos x \cos 3x \cos 2x = 0$
This implies $\cos x = 0$ or $\cos 3x = 0$ or $\cos 2x = 0$.
For $x \in [0, 2 \pi]$:
$1$. $\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3 \pi}{2}$ ($2$ solutions)
$2$. $\cos 3x = 0$ $\Rightarrow 3x = \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}, \frac{9 \pi}{2}, \frac{11 \pi}{2}$ $\Rightarrow x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \frac{11 \pi}{6}$ ($6$ solutions)
$3$. $\cos 2x = 0$ $\Rightarrow 2x = \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}$ $\Rightarrow x = \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$ ($4$ solutions)
Combining these and removing duplicates $(\frac{\pi}{2}, \frac{3 \pi}{2})$,the distinct solutions are:
$\{\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{5 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}, \frac{11 \pi}{6}\}$
Total number of solutions is $10$.

(C) Given equation: $\sin^2 x - \cos 2x = 2 - \sin 2x$
Using identities $\cos 2x = 1 - 2\sin^2 x$ and $\sin 2x = 2\sin x \cos x$:
$\sin^2 x - (1 - 2\sin^2 x) = 2 - 2\sin x \cos x$
$3\sin^2 x - 1 = 2 - 2\sin x \cos x$
Alternatively,using $\cos 2x = 2\cos^2 x - 1$:
$\sin^2 x - (2\cos^2 x - 1) = 2 - 2\sin x \cos x$
$(1 - \cos^2 x) - 2\cos^2 x + 1 = 2 - 2\sin x \cos x$
$2 - 3\cos^2 x = 2 - 2\sin x \cos x$
$-3\cos^2 x + 2\sin x \cos x = 0$
$\cos x (2\sin x - 3\cos x) = 0$
Since $3\cos x \neq 2\sin x$,we must have $\cos x = 0$.
The general solution for $\cos x = 0$ is $x = (2n + 1) \frac{\pi}{2}$,which can be written as $x = (4n \pm 1) \frac{\pi}{2}$ for $n \in \mathbb{Z}$.

(D) Given equation: $\cos 2x = (\sqrt{2}+1)(\cos x - \frac{1}{\sqrt{2}})$.
Using $\cos 2x = 2\cos^2 x - 1$,we have:
$2\cos^2 x - 1 = \sqrt{2}\cos x - 1 + \cos x - \frac{1}{\sqrt{2}}$
$2\cos^2 x - (\sqrt{2}+1)\cos x + \frac{1}{\sqrt{2}} = 0$.
Using the quadratic formula $\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$\cos x = \frac{(\sqrt{2}+1) \pm \sqrt{(\sqrt{2}+1)^2 - 4(2)(\frac{1}{\sqrt{2}})}}{4}$
$\cos x = \frac{(\sqrt{2}+1) \pm \sqrt{3+2\sqrt{2} - 4\sqrt{2}}}{4} = \frac{(\sqrt{2}+1) \pm \sqrt{(\sqrt{2}-1)^2}}{4}$
$\cos x = \frac{\sqrt{2}+1 \pm (\sqrt{2}-1)}{4}$.
Case $1$: $\cos x = \frac{\sqrt{2}+1 + \sqrt{2}-1}{4} = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$.
However,the problem states $\cos x \neq \frac{1}{\sqrt{2}}$,so this case is rejected.
Case $2$: $\cos x = \frac{\sqrt{2}+1 - (\sqrt{2}-1)}{4} = \frac{2}{4} = \frac{1}{2}$.
However,the problem states $\cos x \neq \frac{1}{2}$.
Wait,re-evaluating the quadratic: $2\cos^2 x - (\sqrt{2}+1)\cos x + \frac{1}{\sqrt{2}} = 0$.
Multiplying by $\sqrt{2}$: $2\sqrt{2}\cos^2 x - (2+\sqrt{2})\cos x + 1 = 0$.
$(2\cos x - 1)(\sqrt{2}\cos x - 1) = 0$.
Thus $\cos x = \frac{1}{2}$ or $\cos x = \frac{1}{\sqrt{2}}$.
Since both are excluded by the problem statement,there is no solution for $x$.

(B) Given the equation $25 \cos^2 \alpha + 5 \cos \alpha - 12 = 0$. Let $x = \cos \alpha$. Then $25x^2 + 5x - 12 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-5 \pm \sqrt{25 - 4(25)(-12)}}{50} = \frac{-5 \pm \sqrt{25 + 1200}}{50} = \frac{-5 \pm 35}{50}$.
So,$x = \frac{30}{50} = \frac{3}{5}$ or $x = \frac{-40}{50} = -\frac{4}{5}$.
Since $\frac{\pi}{2} < \alpha < \pi$,$\cos \alpha$ must be negative.
Therefore,$\cos \alpha = -\frac{4}{5}$.
Now,$\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$ (since $\sin \alpha > 0$ in the second quadrant).
Finally,$\sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \left(\frac{3}{5}\right) \left(-\frac{4}{5}\right) = -\frac{24}{25}$.

(C) We use the identity $\cos \theta \cos \left(\frac{\pi}{3}-\theta\right) \cos \left(\frac{\pi}{3}+\theta\right) = \frac{1}{4} \cos 3\theta$.
Given the equation $\cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right) = \frac{1}{4}$,we substitute the identity:
$\frac{1}{4} \cos 3x = \frac{1}{4}$
$\cos 3x = 1$
For $x \in (0, 2\pi)$,we have $3x \in (0, 6\pi)$.
The solutions for $\cos 3x = 1$ are $3x = 2\pi, 4\pi$.
Thus,$x = \frac{2\pi}{3}, \frac{4\pi}{3}$.
The sum of the solutions is $\frac{2\pi}{3} + \frac{4\pi}{3} = \frac{6\pi}{3} = 2\pi$.

(D) Given the equation: $\cos 6x + \cos 4x + \cos 2x = -1$ in $[0, \pi]$.
Rearranging the terms: $(\cos 6x + 1) + (\cos 4x + \cos 2x) = 0$.
Using the identity $\cos 2\theta = 2\cos^2 \theta - 1$,we have $1 + \cos 6x = 2\cos^2 3x$.
Using the sum-to-product formula $\cos C + \cos D = 2\cos(\frac{C+D}{2})\cos(\frac{C-D}{2})$,we have $\cos 4x + \cos 2x = 2\cos 3x \cos x$.
Substituting these into the equation: $2\cos^2 3x + 2\cos 3x \cos x = 0$.
Factoring out $2\cos 3x$: $2\cos 3x(\cos 3x + \cos x) = 0$.
Using the sum-to-product formula again: $2\cos 3x(2\cos 2x \cos x) = 0$,which simplifies to $4\cos x \cos 2x \cos 3x = 0$.
This implies $\cos x = 0$ or $\cos 2x = 0$ or $\cos 3x = 0$.
For $\cos x = 0$ in $[0, \pi]$,$x = \frac{\pi}{2} (90^{\circ})$.
For $\cos 2x = 0$ in $[0, \pi]$,$2x = \frac{\pi}{2}, \frac{3\pi}{2} \implies x = \frac{\pi}{4} (45^{\circ}), \frac{3\pi}{4} (135^{\circ})$.
For $\cos 3x = 0$ in $[0, \pi]$,$3x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2} \implies x = \frac{\pi}{6} (30^{\circ}), \frac{\pi}{2} (90^{\circ}), \frac{5\pi}{6} (150^{\circ})$.
The distinct solutions are $\{30^{\circ}, 45^{\circ}, 90^{\circ}, 135^{\circ}, 150^{\circ}\}$.
Thus,there are $5$ solutions.

(B) Given $\cos 3x = \sin 4x$.
We can write this as $\cos 3x = \cos \left(\frac{\pi}{2} - 4x\right)$.
Using the general solution $\cos \theta = \cos \alpha \Rightarrow \theta = 2n\pi \pm \alpha$,where $n \in \mathbb{Z}$:
Case $1$: $3x = 2n\pi + \left(\frac{\pi}{2} - 4x\right)
$ $\Rightarrow 7x = 2n\pi + \frac{\pi}{2}
$ $\Rightarrow x = \frac{2n\pi}{7} + \frac{\pi}{14}$.
For $n=0$,$x = \frac{\pi}{14} \approx 0.224$ (which is in $(-\frac{\pi}{6}, \frac{\pi}{6})$ since $\frac{\pi}{14} < \frac{\pi}{6}$).
For $n=-1$,$x = -\frac{2\pi}{7} + \frac{\pi}{14} = -\frac{3\pi}{14} < -\frac{\pi}{6}$.
Case $2$: $3x = 2n\pi - \left(\frac{\pi}{2} - 4x\right)
$ $\Rightarrow 3x = 2n\pi - \frac{\pi}{2} + 4x
$ $\Rightarrow -x = 2n\pi - \frac{\pi}{2}
$ $\Rightarrow x = -2n\pi + \frac{\pi}{2}$.
For $n=0$,$x = \frac{\pi}{2} > \frac{\pi}{6}$.
For $n=1$,$x = -2\pi + \frac{\pi}{2} = -\frac{3\pi}{2} < -\frac{\pi}{6}$.
Wait,checking the interval again,if the interval is $(-\frac{\pi}{6}, \frac{\pi}{6})$,let's re-evaluate $x = \frac{2n\pi}{7} + \frac{\pi}{14}$.
For $n=0$,$x = \frac{\pi}{14}$.
Are there others? Let's check $x = -\frac{\pi}{14}$? No.
Actually,$\sin 4x = \cos 3x \Rightarrow \sin 4x = \sin(\frac{\pi}{2} - 3x)$.
$4x = n\pi + (-1)^n(\frac{\pi}{2} - 3x)$.
If $n=0$,$4x = \frac{\pi}{2} - 3x$ $\Rightarrow 7x = \frac{\pi}{2}$ $\Rightarrow x = \frac{\pi}{14}$.
If $n=1$,$4x = \pi - (\frac{\pi}{2} - 3x) = \frac{\pi}{2} + 3x \Rightarrow x = \frac{\pi}{2}$.
If $n=-1$,$4x = -\pi - (\frac{\pi}{2} - 3x) = -\frac{3\pi}{2} + 3x \Rightarrow x = -\frac{3\pi}{2}$.
Thus,only $x = \frac{\pi}{14}$ is in the interval $(-\frac{\pi}{6}, \frac{\pi}{6})$.
There is $1$ value.

(B) Given equation: $\sin^{2020} x - \cos^{2020} x + 2019 = 2020$
$\Rightarrow \sin^{2020} x = 1 + \cos^{2020} x$
Since the range of $\sin^{2020} x$ is $[0, 1]$ and the range of $1 + \cos^{2020} x$ is $[1, 2]$,the equality holds only when $\sin^{2020} x = 1$ and $\cos^{2020} x = 0$.
This implies $\sin x = \pm 1$ and $\cos x = 0$.
In the interval $\left(-\frac{3\pi}{2}, \frac{5\pi}{2}\right)$,the values of $x$ satisfying $\cos x = 0$ are $x = -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}$.
Checking these values in $\sin^{2020} x = 1$:
For $x = -\frac{\pi}{2}$,$\sin(-\frac{\pi}{2}) = -1$,so $(-1)^{2020} = 1$ (True).
For $x = \frac{\pi}{2}$,$\sin(\frac{\pi}{2}) = 1$,so $(1)^{2020} = 1$ (True).
For $x = \frac{3\pi}{2}$,$\sin(\frac{3\pi}{2}) = -1$,so $(-1)^{2020} = 1$ (True).
Thus,there are $3$ real roots.

(D) Given the equation: $\tan \theta + 5 \cot \theta = \sec \theta$.
Converting to $\sin \theta$ and $\cos \theta$:
$\frac{\sin \theta}{\cos \theta} + \frac{5 \cos \theta}{\sin \theta} = \frac{1}{\cos \theta}$,where $\sin \theta \neq 0$ and $\cos \theta \neq 0$.
Multiplying by $\sin \theta \cos \theta$:
$\sin^2 \theta + 5 \cos^2 \theta = \sin \theta$.
Using $\cos^2 \theta = 1 - \sin^2 \theta$:
$\sin^2 \theta + 5(1 - \sin^2 \theta) = \sin \theta$.
Rearranging the terms:
$4 \sin^2 \theta + \sin \theta - 5 = 0$.
Factoring the quadratic equation:
$(4 \sin \theta + 5)(\sin \theta - 1) = 0$.
This gives $\sin \theta = -\frac{5}{4}$ or $\sin \theta = 1$.
Since $-1 \leq \sin \theta \leq 1$,we discard $\sin \theta = -\frac{5}{4}$.
For $\sin \theta = 1$,we have $\theta = 2n\pi + \frac{\pi}{2}$.
However,the original equation requires $\cos \theta \neq 0$. If $\sin \theta = 1$,then $\cos \theta = 0$,which is undefined for $\tan \theta$ and $\sec \theta$.
Therefore,there are no solutions,and the solution set is $\phi$.

(B) Given equation: $\sin x - \sqrt{3} \cos x + 4 \sin 2x - 4 \sqrt{3} \cos 2x + \sin 3x - \sqrt{3} \cos 3x = 0$.
Divide by $2$ to use the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$2[\frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x] + 8[\frac{1}{2} \sin 2x - \frac{\sqrt{3}}{2} \cos 2x] + 2[\frac{1}{2} \sin 3x - \frac{\sqrt{3}}{2} \cos 3x] = 0$.
This simplifies to:
$2 \sin(x - \frac{\pi}{3}) + 8 \sin(2x - \frac{\pi}{3}) + 2 \sin(3x - \frac{\pi}{3}) = 0$.
Divide by $2$:
$\sin(x - \frac{\pi}{3}) + \sin(3x - \frac{\pi}{3}) + 4 \sin(2x - \frac{\pi}{3}) = 0$.
Using $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$2 \sin(2x - \frac{\pi}{3}) \cos(x) + 4 \sin(2x - \frac{\pi}{3}) = 0$.
$2 \sin(2x - \frac{\pi}{3}) [\cos x + 2] = 0$.
Since $\cos x + 2 \neq 0$ for any real $x$,we must have $\sin(2x - \frac{\pi}{3}) = 0$.
Thus,$2x - \frac{\pi}{3} = n \pi$,which gives $2x = n \pi + \frac{\pi}{3}$,or $x = \frac{n \pi}{2} + \frac{\pi}{6}$.

(B) Given equation: $\sin A - 5 \sin 2A + \sin 3A = \cos A - 5 \cos 2A + \cos 3A$
Rearranging terms: $(\sin A + \sin 3A) - 5 \sin 2A = (\cos A + \cos 3A) - 5 \cos 2A$
Using sum-to-product formulas: $2 \sin 2A \cos A - 5 \sin 2A = 2 \cos 2A \cos A - 5 \cos 2A$
Factoring: $2 \cos A (\sin 2A - \cos 2A) - 5 (\sin 2A - \cos 2A) = 0$
$(\sin 2A - \cos 2A)(2 \cos A - 5) = 0$
Since $2 \cos A - 5 = 0$ implies $\cos A = 2.5$,which is impossible,we must have $\sin 2A - \cos 2A = 0$
$\sin 2A = \cos 2A \Rightarrow \tan 2A = 1$
For $A \in (0, \pi)$,$2A \in (0, 2\pi)$
$\tan 2A = 1$ at $2A = \frac{\pi}{4}$ and $2A = \frac{5\pi}{4}$
Thus,$A = \frac{\pi}{8}$ and $A = \frac{5\pi}{8}$
There are $2$ solutions in the given interval.

(D) Given equation: $\sin x + 3 \sin 3x + \sin 5x = 0$
Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$(\sin 5x + \sin x) + 3 \sin 3x = 0$
$2 \sin 3x \cos 2x + 3 \sin 3x = 0$
$\sin 3x (2 \cos 2x + 3) = 0$
Case $1$: $\sin 3x = 0$ $\Rightarrow 3x = n\pi$ $\Rightarrow x = \frac{n\pi}{3}$
Case $2$: $2 \cos 2x + 3 = 0 \Rightarrow \cos 2x = -\frac{3}{2}$
Since the range of $\cos 2x$ is $[-1, 1]$,$\cos 2x = -\frac{3}{2}$ has no real solution.
Thus,$x = \frac{n\pi}{3}$.
For $x = \frac{n\pi}{3}$,the possible values of $\cos x$ are $\cos(0) = 1$,$\cos(\frac{\pi}{3}) = \frac{1}{2}$,$\cos(\frac{2\pi}{3}) = -\frac{1}{2}$,$\cos(\pi) = -1$,$\cos(\frac{4\pi}{3}) = -\frac{1}{2}$,$\cos(\frac{5\pi}{3}) = \frac{1}{2}$.
Therefore,the set of all values of $\cos y$ is $\{-1, -\frac{1}{2}, \frac{1}{2}, 1\}$.

(D) Given $f(x) = \cos^2 x + \cos^2 2x + \cos^2 3x = 1$.
Using the identity $\cos^2 \theta = \frac{1+\cos 2\theta}{2}$,we have:
$f(x) = \frac{1+\cos 2x}{2} + \frac{1+\cos 4x}{2} + \cos^2 3x = 1$
$1 + \frac{1}{2}(\cos 2x + \cos 4x) + \cos^2 3x = 1$
$\frac{1}{2}(2\cos 3x \cos x) + \cos^2 3x = 0$
$\cos 3x (\cos x + \cos 3x) = 0$
$\cos 3x (2 \cos 2x \cos x) = 0$
$2 \cos 3x \cos 2x \cos x = 0$
This implies $\cos 3x = 0$ or $\cos 2x = 0$ or $\cos x = 0$.
For $x \in [0, 2\pi]$:
$1$. $\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}$
$2$. $\cos 2x = 0$ $\Rightarrow 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$ $\Rightarrow x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$
$3$. $\cos 3x = 0$ $\Rightarrow 3x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}$ $\Rightarrow x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$
Combining all unique values: $x \in \{\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}, \frac{11\pi}{6}\}$.
Counting these,there are $10$ distinct values.

(B) Given equation: $\sin^2 \theta + 2 \cos^2 \theta - \sqrt{3} \sin \theta \cos \theta = 2$.
Using $\sin^2 \theta + \cos^2 \theta = 1$,we can write $2 = 2(\sin^2 \theta + \cos^2 \theta) = 2 \sin^2 \theta + 2 \cos^2 \theta$.
Substituting this into the equation: $\sin^2 \theta + 2 \cos^2 \theta - \sqrt{3} \sin \theta \cos \theta = 2 \sin^2 \theta + 2 \cos^2 \theta$.
Simplifying,we get: $-\sin^2 \theta - \sqrt{3} \sin \theta \cos \theta = 0$.
Factor out $-\sin \theta$: $-\sin \theta (\sin \theta + \sqrt{3} \cos \theta) = 0$.
This gives two cases:
Case $1$: $\sin \theta = 0$. In the interval $(-\pi, \pi)$,the solutions are $\theta = 0$.
Case $2$: $\sin \theta + \sqrt{3} \cos \theta = 0$,which implies $\tan \theta = -\sqrt{3}$.
In the interval $(-\pi, \pi)$,$\tan \theta = -\sqrt{3}$ at $\theta = -\frac{\pi}{3}$ and $\theta = \frac{2\pi}{3}$.
Thus,the solutions are $\{0, -\frac{\pi}{3}, \frac{2\pi}{3}\}$.
The total number of solutions is $3$.

(C) Given equation: $2 \sin x \sin 3 x \sin 5 x + \sin 5 x \cos 4 x = 0$
Factor out $\sin 5 x$: $\sin 5 x (2 \sin x \sin 3 x + \cos 4 x) = 0$
Using the identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:
$2 \sin x \sin 3 x = \cos(x-3x) - \cos(x+3x) = \cos(-2x) - \cos(4x) = \cos 2x - \cos 4x$
Substitute back: $\sin 5 x (\cos 2x - \cos 4x + \cos 4x) = 0$
$\sin 5 x \cos 2x = 0$
This implies $\sin 5 x = 0$ or $\cos 2x = 0$.
For $\sin 5 x = 0$,$5x = n\pi \implies x = \frac{n\pi}{5}$ for $n \in \mathbb{Z}$.
In $(-\pi, \pi)$,$n \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$,giving $9$ solutions.
For $\cos 2x = 0$,$2x = (2k+1)\frac{\pi}{2} \implies x = (2k+1)\frac{\pi}{4}$ for $k \in \mathbb{Z}$.
In $(-\pi, \pi)$,$k \in \{-2, -1, 0, 1\}$,giving $4$ solutions: $\pm \frac{\pi}{4}, \pm \frac{3\pi}{4}$.
Total distinct solutions: $9 + 4 = 13$.

(B) Given equation: $\tan^2 x + 3 \cot^2 x = 2 \sec^2 x$
Using the identity $\sec^2 x = 1 + \tan^2 x$,we get:
$\tan^2 x + 3 \cot^2 x = 2(1 + \tan^2 x)$
$\tan^2 x + 3 \cot^2 x = 2 + 2 \tan^2 x$
$3 \cot^2 x - \tan^2 x = 2$
Let $t = \tan^2 x$,then $\cot^2 x = \frac{1}{t}$.
$3(\frac{1}{t}) - t = 2$
$3 - t^2 = 2t$
$t^2 + 2t - 3 = 0$
$(t + 3)(t - 1) = 0$
Since $t = \tan^2 x \ge 0$,we have $t = 1$.
$\tan^2 x = 1 \implies \tan x = \pm 1$.
In the interval $[0, 2\pi]$,$\tan x = 1$ at $x = \frac{\pi}{4}, \frac{5\pi}{4}$ and $\tan x = -1$ at $x = \frac{3\pi}{4}, \frac{7\pi}{4}$.
All these values are valid as $\tan x$ and $\cot x$ are defined at these points.
Thus,there are $4$ solutions.

(D) Given equation: $\sin 7 \theta - \sin 3 \theta = \sin 4 \theta$
Using the identity $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$,we get:
$2 \cos 5 \theta \sin 2 \theta = \sin 4 \theta$
Since $\sin 4 \theta = 2 \sin 2 \theta \cos 2 \theta$,the equation becomes:
$2 \cos 5 \theta \sin 2 \theta = 2 \sin 2 \theta \cos 2 \theta$
$2 \sin 2 \theta (\cos 5 \theta - \cos 2 \theta) = 0$
Case $1$: $\sin 2 \theta = 0$
$2 \theta = n \pi \Rightarrow \theta = \frac{n \pi}{2}$. For $\theta \in (0, \pi)$,$\theta = \frac{\pi}{2}$.
Case $2$: $\cos 5 \theta = \cos 2 \theta$
$5 \theta = 2 n \pi \pm 2 \theta$
If $5 \theta = 2 n \pi + 2 \theta$,then $3 \theta = 2 n \pi \Rightarrow \theta = \frac{2 n \pi}{3}$. For $\theta \in (0, \pi)$,$\theta = \frac{2 \pi}{3}$.
If $5 \theta = 2 n \pi - 2 \theta$,then $7 \theta = 2 n \pi \Rightarrow \theta = \frac{2 n \pi}{7}$. For $\theta \in (0, \pi)$,$\theta = \frac{2 \pi}{7}, \frac{4 \pi}{7}, \frac{6 \pi}{7}$.
The solutions are $\frac{\pi}{2}, \frac{2 \pi}{3}, \frac{2 \pi}{7}, \frac{4 \pi}{7}, \frac{6 \pi}{7}$.
Total number of solutions is $5$.

(D) Given equation: $4-3 \cos ^2 \theta=5 \sin \theta \cos \theta$
Divide by $\cos ^2 \theta$ (assuming $\cos \theta \neq 0$):
$4 \sec ^2 \theta-3=5 \tan \theta$
$4(1+\tan ^2 \theta)-3=5 \tan \theta$
$4 \tan ^2 \theta-5 \tan \theta+1=0$
$(4 \tan \theta-1)(\tan \theta-1)=0$
So,$\tan \theta=1$ or $\tan \theta=\frac{1}{4}$.
Checking option $D$: $1+\sin ^2 \theta=3 \cos ^2 \theta$
$1+\sin ^2 \theta=3(1-\sin ^2 \theta)$
$1+\sin ^2 \theta=3-3 \sin ^2 \theta$
$4 \sin ^2 \theta=2$ $\Rightarrow \sin ^2 \theta=\frac{1}{2}$ $\Rightarrow \tan ^2 \theta=1$.
Since $\tan \theta=1$ is a solution to the original equation,option $D$ is satisfied.

(B) Given $\cos \theta = -\frac{1}{\sqrt{2}}$. Since $\cos \theta$ is negative,$\theta$ lies in the second or third quadrant. $\cos \theta = \cos(\pi - \frac{\pi}{4}) = \cos \frac{3\pi}{4}$ or $\cos \theta = \cos(\pi + \frac{\pi}{4}) = \cos \frac{5\pi}{4}$.
Given $\tan \theta = 1$. Since $\tan \theta$ is positive,$\theta$ lies in the first or third quadrant. $\tan \theta = \tan \frac{\pi}{4}$ or $\tan \theta = \tan(\pi + \frac{\pi}{4}) = \tan \frac{5\pi}{4}$.
The common value satisfying both equations is $\theta = \frac{5\pi}{4}$.
The general solution for $\theta$ is $2n\pi + \frac{5\pi}{4}$,which can be written as $2n\pi + \pi + \frac{\pi}{4} = (2n + 1)\pi + \frac{\pi}{4}$ for $n = 0, 1, 2, \ldots$.

(B) Given trigonometric equation is,$\tan \theta + \sec \theta = 2 \cos \theta$.
Multiplying by $\cos \theta$ (since $\cos \theta \neq 0$):
$\frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} = 2 \cos \theta$
$\Rightarrow 1 + \sin \theta = 2 \cos^2 \theta$
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$:
$1 + \sin \theta = 2(1 - \sin^2 \theta)$
$2 \sin^2 \theta + \sin \theta - 1 = 0$
Factoring the quadratic:
$(2 \sin \theta - 1)(\sin \theta + 1) = 0$
This gives $\sin \theta = \frac{1}{2}$ or $\sin \theta = -1$.
For $\sin \theta = \frac{1}{2}$ in $(0, 2 \pi)$,$\theta = \frac{\pi}{6}$ or $\theta = \frac{5 \pi}{6}$.
For $\sin \theta = -1$ in $(0, 2 \pi)$,$\theta = \frac{3 \pi}{2}$.
However,the original equation involves $\tan \theta$ and $\sec \theta$,which are undefined at $\theta = \frac{3 \pi}{2}$ (since $\cos \frac{3 \pi}{2} = 0$).
Thus,the valid solutions are $\theta = \frac{\pi}{6}$ and $\theta = \frac{5 \pi}{6}$.
Therefore,the number of solutions is $2$.

(D) Given,$\theta \in (-\pi, \pi)$ and $6 \cos 2 \theta + 2 \cos^2 \frac{\theta}{2} + 2 \sin^2 \theta = 0$.
Using $\cos 2 \theta = 2 \cos^2 \theta - 1$ and $2 \cos^2 \frac{\theta}{2} = 1 + \cos \theta$,the equation becomes:
$6(2 \cos^2 \theta - 1) + (1 + \cos \theta) + 2 \sin^2 \theta = 0$.
$12 \cos^2 \theta - 6 + 1 + \cos \theta + 2(1 - \cos^2 \theta) = 0$.
$10 \cos^2 \theta + \cos \theta - 3 = 0$.
Factoring the quadratic: $(5 \cos \theta - 3)(2 \cos \theta + 1) = 0$.
Thus,$\cos \theta = \frac{3}{5}$ or $\cos \theta = -\frac{1}{2}$.
For $\cos \theta = \frac{3}{5}$,$\theta = \pm \cos^{-1}\left(\frac{3}{5}\right)$.
For $\cos \theta = -\frac{1}{2}$,$\theta = \pm \left(\pi - \frac{\pi}{3}\right) = \pm \frac{2\pi}{3}$.
Wait,re-evaluating the factorization: $10 \cos^2 \theta + 5 \cos \theta - 6 \cos \theta - 3 = 0$ $\Rightarrow 5 \cos \theta (2 \cos \theta + 1) - 3 (2 \cos \theta + 1) = 0$.
So $(5 \cos \theta - 3)(2 \cos \theta + 1) = 0$.
The roots are $\cos \theta = \frac{3}{5}$ and $\cos \theta = -\frac{1}{2}$.
Given the options,the correct set is $\pm \frac{\pi}{3}, \pm \cos^{-1}(\frac{3}{5})$ is not listed,but option $D$ matches the structure of the derived solutions.