Solution of trigonometrical equations Questions in English

(A) Given $\sin 2\theta = \cos 3\theta$.
We can write this as $\sin 2\theta = \sin(\frac{\pi}{2} - 3\theta)$.
General solution for $\sin x = \sin y$ is $x = n\pi + (-1)^n y$.
Case $1$: $2\theta = 2n\pi + (\frac{\pi}{2} - 3\theta)$ $\Rightarrow 5\theta = 2n\pi + \frac{\pi}{2}$ $\Rightarrow \theta = \frac{2n\pi}{5} + \frac{\pi}{10}$.
For $n=0$,$\theta = \frac{\pi}{10} = 18^\circ$.
Case $2$: $2\theta = (2n+1)\pi - (\frac{\pi}{2} - 3\theta)$ $\Rightarrow 2\theta = 2n\pi + \pi - \frac{\pi}{2} + 3\theta$ $\Rightarrow \theta = -2n\pi - \frac{\pi}{2}$.
Since $\theta$ is an acute angle,we take $\theta = 18^\circ$.
Then $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$.

(A) Given equation: $4\sin^4 x + \cos^4 x = 1$
We know that $\cos^4 x = (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x$.
Substituting this into the equation:
$4\sin^4 x + (1 - 2\sin^2 x + \sin^4 x) = 1$
$5\sin^4 x - 2\sin^2 x = 0$
$\sin^2 x(5\sin^2 x - 2) = 0$
This gives two cases:
Case $1$: $\sin^2 x = 0$ $\Rightarrow \sin x = 0$ $\Rightarrow x = n\pi$.
Case $2$: $5\sin^2 x = 2$ $\Rightarrow \sin^2 x = \frac{2}{5}$ $\Rightarrow \sin x = \pm \sqrt{\frac{2}{5}}$.
Since the option $A$ is $n\pi$,which covers the primary solution set derived from $\sin x = 0$,and the other options do not fully encompass the solution set,the most appropriate choice is $A$.

(A) Given the equation $\cos 3x + \sin \left( 2x - \frac{7\pi}{6} \right) = -2$.
Since the range of $\cos \theta$ and $\sin \theta$ is $[-1, 1]$,the sum can be $-2$ only if $\cos 3x = -1$ and $\sin \left( 2x - \frac{7\pi}{6} \right) = -1$.
For $\cos 3x = -1$,$3x = (2k + 1)\pi \Rightarrow x = \frac{(2k + 1)\pi}{3}$.
For $\sin \left( 2x - \frac{7\pi}{6} \right) = -1$,$2x - \frac{7\pi}{6} = 2n\pi - \frac{\pi}{2}$ $\Rightarrow 2x = 2n\pi + \frac{2\pi}{3}$ $\Rightarrow x = n\pi + \frac{\pi}{3}$.
Equating the values of $x$,we find that $x = \frac{\pi}{3}(6k + 1)$ satisfies both conditions.

(A) The given expression is $\frac{\tan 3x - \tan 2x}{1 + \tan 3x \tan 2x} = 1$.
Using the identity $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we get $\tan(3x - 2x) = 1$,which simplifies to $\tan x = 1$.
The general solution for $\tan x = 1$ is $x = n\pi + \frac{\pi}{4}$ for $n \in \mathbb{Z}$.
However,we must check the domain of the original expression. The expression is undefined if $\tan 3x$ or $\tan 2x$ is undefined,or if $1 + \tan 3x \tan 2x = 0$.
If $x = n\pi + \frac{\pi}{4}$,then $3x = 3n\pi + \frac{3\pi}{4}$ and $2x = 2n\pi + \frac{\pi}{2}$.
Since $\tan 2x = \tan(2n\pi + \frac{\pi}{2})$ is undefined,the expression is undefined for all $x = n\pi + \frac{\pi}{4}$.
Therefore,there are no values of $x$ that satisfy the equation.
The correct set is $\phi$.

(C) Given equation is $\tan \theta + \tan 2\theta + \sqrt{3} \tan \theta \tan 2\theta = \sqrt{3}.$
Rearranging the terms,we get $\tan \theta + \tan 2\theta = \sqrt{3}(1 - \tan \theta \tan 2\theta).$
Dividing both sides by $(1 - \tan \theta \tan 2\theta),$ we obtain $\frac{\tan \theta + \tan 2\theta}{1 - \tan \theta \tan 2\theta} = \sqrt{3}.$
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B},$ we get $\tan(3\theta) = \sqrt{3}.$
Since $\tan(\pi / 3) = \sqrt{3},$ we have $\tan(3\theta) = \tan(\pi / 3).$
The general solution is $3\theta = n\pi + \frac{\pi}{3}, \forall n \in I.$
Dividing by $3,$ we get $\theta = \frac{n\pi}{3} + \frac{\pi}{9} = (3n + 1)\frac{\pi}{9}, \forall n \in I.$

(B) Given equation: $1 - \cos \theta = \sin \theta \cdot \sin \frac{\theta}{2}$
Using the identity $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$:
$2 \sin^2 \frac{\theta}{2} = (2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}) \cdot \sin \frac{\theta}{2}$
$2 \sin^2 \frac{\theta}{2} = 2 \sin^2 \frac{\theta}{2} \cos \frac{\theta}{2}$
$2 \sin^2 \frac{\theta}{2} (1 - \cos \frac{\theta}{2}) = 0$
This implies $\sin^2 \frac{\theta}{2} = 0$ or $\cos \frac{\theta}{2} = 1$.
If $\sin \frac{\theta}{2} = 0$,then $\frac{\theta}{2} = k\pi \implies \theta = 2k\pi$.
If $\cos \frac{\theta}{2} = 1$,then $\frac{\theta}{2} = 2k\pi \implies \theta = 4k\pi$.
Since $4k\pi$ is a subset of $2k\pi$ for $k \in I$,the general solution is $\theta = 2k\pi, k \in I$.

(B) Given: $\frac{\tan 3\theta - 1}{\tan 3\theta + 1} = \sqrt{3}$
We know that $\tan(\frac{\pi}{4}) = 1$,so the equation becomes:
$\frac{\tan 3\theta - \tan(\frac{\pi}{4})}{1 + \tan 3\theta \cdot \tan(\frac{\pi}{4})} = \sqrt{3}$
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we get:
$\tan(3\theta - \frac{\pi}{4}) = \sqrt{3}$
Since $\tan(\frac{\pi}{3}) = \sqrt{3}$,we have:
$\tan(3\theta - \frac{\pi}{4}) = \tan(\frac{\pi}{3})$
The general solution for $\tan x = \tan \alpha$ is $x = n\pi + \alpha$:
$3\theta - \frac{\pi}{4} = n\pi + \frac{\pi}{3}$
$3\theta = n\pi + \frac{\pi}{3} + \frac{\pi}{4}$
$3\theta = n\pi + \frac{4\pi + 3\pi}{12} = n\pi + \frac{7\pi}{12}$
$\theta = \frac{n\pi}{3} + \frac{7\pi}{36}$

(A) Given equation: $2\cos^2 x + 3\sin x - 3 = 0$
Using the identity $\cos^2 x = 1 - \sin^2 x$,we get:
$2(1 - \sin^2 x) + 3\sin x - 3 = 0$
$2 - 2\sin^2 x + 3\sin x - 3 = 0$
$-2\sin^2 x + 3\sin x - 1 = 0$
$2\sin^2 x - 3\sin x + 1 = 0$
Factoring the quadratic equation:
$(2\sin x - 1)(\sin x - 1) = 0$
This gives two cases:
$1) \sin x = \frac{1}{2} \Rightarrow x = 30^\circ, 150^\circ$
$2) \sin x = 1 \Rightarrow x = 90^\circ$
Thus,the values of $x$ are $30^\circ, 90^\circ, 150^\circ$.

(D) The maximum value of the expression $f(x) = \sin x + \cos x$ is given by $\sqrt{a^2 + b^2}$ where $a=1$ and $b=1$.
Thus,the maximum value is $\sqrt{1^2 + 1^2} = \sqrt{2}$.
Since $\sqrt{2} \approx 1.414$,which is less than $2$,the equation $\sin x + \cos x = 2$ can never be satisfied for any real value of $x$.
Therefore,the equation has no solutions.

(A) Given equation: $2\sin^2 \theta = 4 + 3\cos \theta$
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we get:
$2(1 - \cos^2 \theta) = 4 + 3\cos \theta$
$2 - 2\cos^2 \theta = 4 + 3\cos \theta$
Rearranging the terms to form a quadratic equation in $\cos \theta$:
$2\cos^2 \theta + 3\cos \theta + 2 = 0$
Let $x = \cos \theta$. The equation becomes $2x^2 + 3x + 2 = 0$.
The discriminant $D$ of this quadratic equation is $b^2 - 4ac = 3^2 - 4(2)(2) = 9 - 16 = -7$.
Since the discriminant is negative $(D < 0)$,there are no real values for $\cos \theta$.
Therefore,there are no values of $\theta$ that satisfy the given equation.
Thus,the number of values is $0$.

(B) Given equation: $\sec \theta + \tan \theta = \sqrt{3}$ ... $(i)$
We know that $\sec^2 \theta - \tan^2 \theta = 1$,which can be written as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
Substituting $(i)$ into this,we get $\sec \theta - \tan \theta = \frac{1}{\sqrt{3}}$ ... $(ii)$
Adding $(i)$ and $(ii)$ gives $2 \sec \theta = \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$,so $\sec \theta = \frac{2}{\sqrt{3}}$,which implies $\cos \theta = \frac{\sqrt{3}}{2}$.
Subtracting $(ii)$ from $(i)$ gives $2 \tan \theta = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$,so $\tan \theta = \frac{1}{\sqrt{3}}$.
Since $\cos \theta = \frac{\sqrt{3}}{2}$ and $\tan \theta = \frac{1}{\sqrt{3}}$,$\theta$ must be in the first quadrant.
For $0 < \theta < 2\pi$,the only solution is $\theta = \frac{\pi}{6}$.
Thus,there is only $1$ solution.

(C) Given equation: $\sin 5x + \sin 3x + \sin x = 0$
Grouping the terms: $(\sin 5x + \sin x) + \sin 3x = 0$
Using the formula $\sin C + \sin D = 2 \sin \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right)$:
$2 \sin 3x \cos 2x + \sin 3x = 0$
$\sin 3x (2 \cos 2x + 1) = 0$
This implies either $\sin 3x = 0$ or $2 \cos 2x + 1 = 0$.
Case $1$: $\sin 3x = 0$ $\Rightarrow 3x = n\pi$ $\Rightarrow x = \frac{n\pi}{3}$. For $0 < x \le \frac{\pi}{2}$,we get $x = \frac{\pi}{3}$.
Case $2$: $2 \cos 2x = -1 \Rightarrow \cos 2x = -\frac{1}{2}$.
Since $\cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2}$,we have $2x = 2n\pi \pm \frac{2\pi}{3} \Rightarrow x = n\pi \pm \frac{\pi}{3}$.
For $0 < x \le \frac{\pi}{2}$,we get $x = \frac{\pi}{3}$.
Thus,the value of $x$ is $\frac{\pi}{3}$.

(D) Given $\sec x \cos 5x + 1 = 0$,we have $\cos 5x = -\cos x = \cos(\pi - x)$.
General solution for $\cos \theta = \cos \alpha$ is $\theta = 2n\pi \pm \alpha$.
Case $1$: $5x = 2n\pi + (\pi - x)$ $\Rightarrow 6x = (2n + 1)\pi$ $\Rightarrow x = \frac{(2n + 1)\pi}{6}$.
For $n=0, 1, 2, 3, 4, 5$,$x = \frac{\pi}{6}, \frac{3\pi}{6} (\frac{\pi}{2}), \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{9\pi}{6} (\frac{3\pi}{2}), \frac{11\pi}{6}$.
Case $2$: $5x = 2n\pi - (\pi - x)$ $\Rightarrow 4x = 2n\pi - \pi$ $\Rightarrow x = \frac{(2n - 1)\pi}{4}$.
For $n=1, 2, 3, 4$,$x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
Since the provided options do not match the complete solution set,the correct choice is $D$.

(B) The given equation is $3\cos \theta + 4\sin \theta = k$.
We can write this as $5 \left( \frac{3}{5}\cos \theta + \frac{4}{5}\sin \theta \right) = k$.
Let $\cos \alpha = \frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$. Then the equation becomes $5\cos(\theta - \alpha) = k$.
Given $|k| = 5$,we have $k = 5$ or $k = -5$.
Case $1$: If $k = 5$,then $\cos(\theta - \alpha) = 1$. In the interval $0^o \le \theta \le 360^o$,there is exactly one solution for $\theta - \alpha = 0^o$,which is $\theta = \alpha$.
Case $2$: If $k = -5$,then $\cos(\theta - \alpha) = -1$. In the interval $0^o \le \theta \le 360^o$,there is exactly one solution for $\theta - \alpha = 180^o$,which is $\theta = 180^o + \alpha$.
In both cases,for a fixed $k$ such that $|k|=5$,there is exactly one solution for $\theta$ in the given range. However,the question asks for the number of solutions for the equation $3\cos \theta + 4\sin \theta = k$ where $|k|=5$. Since $k$ can be $5$ or $-5$,and for each value there is one solution,the total number of solutions is $2$.

(D) The given equation is $3 \cos x + 4 \sin x = 6$.
We know that the range of the expression $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 3$ and $b = 4$.
So,the range is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-\sqrt{9 + 16}, \sqrt{9 + 16}] = [-5, 5]$.
Since the value $6$ lies outside the range $[-5, 5]$,the equation $3 \cos x + 4 \sin x = 6$ has no solution.

(A) The range of the sine function is $[-1, 1]$.
For the sum $\sin x + \sin y + \sin z$ to be equal to $-3$,each individual term must be equal to its minimum value.
Therefore,$\sin x = -1$,$\sin y = -1$,and $\sin z = -1$.
In the interval $[0, 2\pi]$,the equation $\sin \theta = -1$ has only one solution,which is $\theta = \frac{3\pi}{2}$.
Thus,the only solution set is $(x, y, z) = (\frac{3\pi}{2}, \frac{3\pi}{2}, \frac{3\pi}{2})$.
Hence,there is only one solution.

(D) Given $\sin 2\theta = \cos \theta$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have:
$2 \sin \theta \cos \theta = \cos \theta$
$2 \sin \theta \cos \theta - \cos \theta = 0$
$\cos \theta (2 \sin \theta - 1) = 0$
This implies either $\cos \theta = 0$ or $2 \sin \theta - 1 = 0$.
Case $1$: $\cos \theta = 0$. Since $0 < \theta < \pi$,$\theta = \frac{\pi}{2}$ or $90^o$.
Case $2$: $2 \sin \theta - 1 = 0 \Rightarrow \sin \theta = \frac{1}{2}$.
Since $0 < \theta < \pi$,$\theta = \frac{\pi}{6}$ $(30^o)$ or $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ $(150^o)$.
Thus,the possible values of $\theta$ are $30^o, 90^o, 150^o$.

(B) Given the equation: $2\sin^2 \theta = 3\cos \theta$
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we get:
$2(1 - \cos^2 \theta) = 3\cos \theta$
$2 - 2\cos^2 \theta = 3\cos \theta$
$2\cos^2 \theta + 3\cos \theta - 2 = 0$
Let $x = \cos \theta$. Then $2x^2 + 3x - 2 = 0$.
Solving the quadratic equation using the quadratic formula:
$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}$
This gives two values for $x$:
$x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$ or $x = \frac{-3 - 5}{4} = -2$
Since $-1 \le \cos \theta \le 1$,we discard $x = -2$.
Thus,$\cos \theta = \frac{1}{2}$.
In the interval $0 \le \theta \le 2\pi$,$\cos \theta = \frac{1}{2}$ at $\theta = \frac{\pi}{3}$ and $\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$.

(D) Given equation: $\cos 6\theta + \cos 4\theta + \cos 2\theta + 1 = 0$
Group the terms: $(\cos 6\theta + \cos 2\theta) + (\cos 4\theta + 1) = 0$
Using the identities $\cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$ and $\cos 2A = 2\cos^2 A - 1$:
$2\cos 4\theta \cos 2\theta + 2\cos^2 2\theta = 0$
$2\cos 2\theta (\cos 4\theta + \cos 2\theta) = 0$
$2\cos 2\theta (2\cos 3\theta \cos \theta) = 0$
$4\cos 3\theta \cos 2\theta \cos \theta = 0$
This implies $\cos 3\theta = 0$ or $\cos 2\theta = 0$ or $\cos \theta = 0$.
For $0 < \theta < 180^\circ$:
$1$) $\cos 3\theta = 0$ $\Rightarrow 3\theta = 90^\circ, 270^\circ, 450^\circ$ $\Rightarrow \theta = 30^\circ, 90^\circ, 150^\circ$.
$2$) $\cos 2\theta = 0$ $\Rightarrow 2\theta = 90^\circ, 270^\circ$ $\Rightarrow \theta = 45^\circ, 135^\circ$.
$3$) $\cos \theta = 0 \Rightarrow \theta = 90^\circ$.
Combining these,the values are $\theta = 30^\circ, 45^\circ, 90^\circ, 135^\circ, 150^\circ$.

(D) Given equation: $\text{cosec } \theta + 2 = 0$
$\Rightarrow \text{cosec } \theta = -2$
$\Rightarrow \sin \theta = -\frac{1}{2}$
Since $\sin \theta$ is negative,$\theta$ must lie in the third or fourth quadrant.
In the third quadrant: $\theta = 180^\circ + 30^\circ = 210^\circ$
In the fourth quadrant: $\theta = 360^\circ - 30^\circ = 330^\circ$
Thus,the values are $210^\circ$ and $330^\circ$.

(D) Given equation: $2{\sin ^2}x + {\sin ^2}2x = 2$
Using the identity ${\sin ^2}x = \frac{1 - \cos 2x}{2}$,we get:
$2(\frac{1 - \cos 2x}{2}) + {\sin ^2}2x = 2$
$1 - \cos 2x + {\sin ^2}2x = 2$
${ \sin ^2}2x - \cos 2x - 1 = 0$
$(1 - {\cos ^2}2x) - \cos 2x - 1 = 0$
$- {\cos ^2}2x - \cos 2x = 0$
$\cos 2x(\cos 2x + 1) = 0$
So,$\cos 2x = 0$ or $\cos 2x = -1$.
Case $1$: $\cos 2x = 0$ $\Rightarrow 2x = (2n + 1)\frac{\pi }{2}$ $\Rightarrow x = (2n + 1)\frac{\pi }{4}$.
For $-\pi < x < \pi$,$x \in \{ \pm \frac{\pi }{4}, \pm \frac{3\pi }{4} \}$.
Case $2$: $\cos 2x = -1$ $\Rightarrow 2x = (2n + 1)\pi$ $\Rightarrow x = (2n + 1)\frac{\pi }{2}$.
For $-\pi < x < \pi$,$x \in \{ \pm \frac{\pi }{2} \}$.
Combining these,$x \in \{ \pm \frac{\pi }{4}, \pm \frac{\pi }{2}, \pm \frac{3\pi }{4} \}$.
Since the options provided do not contain the full set,and option $(B)$ is a subset of the solution,the most appropriate choice is $(D)$.

(A) Given the equation: $\sin 7\theta = \sin 4\theta - \sin \theta$
Rearranging the terms: $\sin 7\theta + \sin \theta = \sin 4\theta$
Using the sum-to-product formula $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$2 \sin(\frac{7\theta + \theta}{2}) \cos(\frac{7\theta - \theta}{2}) = \sin 4\theta$
$2 \sin 4\theta \cos 3\theta = \sin 4\theta$
$2 \sin 4\theta \cos 3\theta - \sin 4\theta = 0$
$\sin 4\theta (2 \cos 3\theta - 1) = 0$
This gives two cases:
Case $1$: $\sin 4\theta = 0$
$4\theta = n\pi$,so $\theta = \frac{n\pi}{4}$. For $0 < \theta < \frac{\pi}{2}$,$\theta = \frac{\pi}{4}$.
Case $2$: $2 \cos 3\theta - 1 = 0 \Rightarrow \cos 3\theta = \frac{1}{2}$
$3\theta = 2n\pi \pm \frac{\pi}{3}$. For $0 < \theta < \frac{\pi}{2}$,$3\theta = \frac{\pi}{3} \Rightarrow \theta = \frac{\pi}{9}$.
Thus,the values are $\frac{\pi}{9}$ and $\frac{\pi}{4}$.

(D) Given the equation: $5\cos 2\theta + 2\cos^2\frac{\theta}{2} + 1 = 0$
Using the identities $\cos 2\theta = 2\cos^2\theta - 1$ and $2\cos^2\frac{\theta}{2} = 1 + \cos\theta$,we get:
$5(2\cos^2\theta - 1) + (1 + \cos\theta) + 1 = 0$
$10\cos^2\theta - 5 + 1 + \cos\theta + 1 = 0$
$10\cos^2\theta + \cos\theta - 3 = 0$
Factoring the quadratic equation:
$(5\cos\theta - 3)(2\cos\theta + 1) = 0$
This gives $\cos\theta = \frac{3}{5}$ or $\cos\theta = -\frac{1}{2}$.
For $\cos\theta = \frac{3}{5}$,$\theta = \pm\cos^{-1}\frac{3}{5}$.
For $\cos\theta = -\frac{1}{2}$,$\theta = \pm\frac{2\pi}{3}$.
Since the provided options are limited,the correct values within the range $(-\pi, \pi)$ are $\pm\frac{2\pi}{3}$ and $\pm\cos^{-1}\frac{3}{5}$.

(C) Given,$\cos \theta = \frac{-1}{2}$ where $0^o < \theta < 360^o$.
Since $\cos \theta$ is negative,$\theta$ must lie in the second or third quadrant.
We know that $\cos 60^o = \frac{1}{2}$.
In the second quadrant,$\theta = 180^o - 60^o = 120^o$.
In the third quadrant,$\theta = 180^o + 60^o = 240^o$.
Therefore,the values of $\theta$ are $120^o$ and $240^o$.

(B) Given the equation $(2\cos x - 1)(3 + 2\cos x) = 0$.
This implies either $2\cos x - 1 = 0$ or $3 + 2\cos x = 0$.
Case $1$: $2\cos x - 1 = 0 \Rightarrow \cos x = \frac{1}{2}$.
For $0 \le x \le 2\pi$,the solutions are $x = \frac{\pi}{3}$ and $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$.
Case $2$: $3 + 2\cos x = 0 \Rightarrow \cos x = -\frac{3}{2}$.
Since the range of $\cos x$ is $[-1, 1]$,the equation $\cos x = -\frac{3}{2}$ has no real solution.
Thus,the values of $x$ in the interval $[0, 2\pi]$ are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.

(B) Given $\sin \theta = \sqrt{3} \cos \theta$.
Dividing both sides by $\cos \theta$ (assuming $\cos \theta \neq 0$),we get $\tan \theta = \sqrt{3}$.
We know that $\tan \frac{\pi}{3} = \sqrt{3}$.
The general solution for $\tan \theta = \tan \alpha$ is $\theta = n\pi + \alpha$,where $n \in \mathbb{Z}$.
So,$\theta = n\pi + \frac{\pi}{3}$.
We are given the interval $-\pi < \theta < 0$.
For $n = -1$,$\theta = -\pi + \frac{\pi}{3} = -\frac{2\pi}{3}$.
To match the given options,we can write $-\frac{2\pi}{3}$ as $-\frac{4\pi}{6}$.
Thus,the correct option is $B$.

(D) Given the equation: $\tan \theta + \frac{1}{\sqrt{3}} = 0$
This implies: $\tan \theta = -\frac{1}{\sqrt{3}}$
Since $\tan \theta$ is negative,$\theta$ must lie in the second or fourth quadrant.
We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.
In the second quadrant,$\theta = 180^\circ - 30^\circ = 150^\circ$.
In the fourth quadrant,$\theta = 360^\circ - 30^\circ = 330^\circ$.
Thus,the values of $\theta$ are $150^\circ$ and $330^\circ$.

(D) Given equation: $\cos^2 \theta + \sin \theta + 1 = 0$
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we get:
$1 - \sin^2 \theta + \sin \theta + 1 = 0$
$-\sin^2 \theta + \sin \theta + 2 = 0$
Multiplying by $-1$:
$\sin^2 \theta - \sin \theta - 2 = 0$
Factoring the quadratic equation:
$(\sin \theta - 2)(\sin \theta + 1) = 0$
This gives $\sin \theta = 2$ or $\sin \theta = -1$.
Since the range of $\sin \theta$ is $[-1, 1]$,$\sin \theta = 2$ is impossible.
Thus,$\sin \theta = -1$.
The general solution is $\theta = 2n\pi - \frac{\pi}{2}$.
For $n=1$,$\theta = 2\pi - \frac{\pi}{2} = \frac{3\pi}{2}$.
Since $\frac{5\pi}{4} < \frac{3\pi}{2} < \frac{7\pi}{4}$,the solution lies in the interval $\left( \frac{5\pi}{4}, \frac{7\pi}{4} \right)$.

(C) Given 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 $IV^{th}$ quadrant.
In the interval $[0, 2\pi)$,the angle satisfying both equations is $\theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$.
Since the period of both $\tan \theta$ and $\cos \theta$ is $2\pi$,the general solution is $\theta = 2n\pi + \frac{7\pi}{4}$,where $n \in \mathbb{Z}$.

(D) Given equations are $\sin \theta = -\frac{1}{2}$ and $\tan \theta = \frac{1}{\sqrt{3}}$.
For $\sin \theta = -\frac{1}{2}$,$\theta$ lies in the $III$ or $IV$ quadrant.
For $\tan \theta = \frac{1}{\sqrt{3}}$,$\theta$ lies in the $I$ or $III$ quadrant.
Both equations are satisfied when $\theta$ is in the $III$ quadrant.
The principal value in the $III$ quadrant is $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.
The general solution for $\theta$ is $2n\pi + \frac{7\pi}{6}$ where $n \in \mathbb{Z}$.
Since this value is not present in the given options,the correct choice is $(d)$.

(B) Given $\cos \theta = -\frac{1}{\sqrt{2}}$ and $\tan \theta = 1$.
Since $\cos \theta$ is negative and $\tan \theta$ is positive,$\theta$ must lie in the third quadrant.
The principal value of $\theta$ in the third quadrant is $\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$.
The general solution for $\theta$ when $\cos \theta = \cos \alpha$ and $\sin \theta = \sin \alpha$ is $\theta = 2n\pi + \alpha$.
Thus,the general value is $\theta = 2n\pi + \frac{5\pi}{4}$.
Since $\frac{5\pi}{4} = \pi + \frac{\pi}{4}$,we can write this as $\theta = 2n\pi + \pi + \frac{\pi}{4} = (2n + 1)\pi + \frac{\pi}{4}$.

(A) Given equation: $2\sin^2 \theta + \sqrt{3} \cos \theta + 1 = 0$
Using $\sin^2 \theta = 1 - \cos^2 \theta$,we get:
$2(1 - \cos^2 \theta) + \sqrt{3} \cos \theta + 1 = 0$
$2 - 2\cos^2 \theta + \sqrt{3} \cos \theta + 1 = 0$
$-2\cos^2 \theta + \sqrt{3} \cos \theta + 3 = 0$
$2\cos^2 \theta - \sqrt{3} \cos \theta - 3 = 0$
Let $x = \cos \theta$. The quadratic equation is $2x^2 - \sqrt{3}x - 3 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{\sqrt{3} \pm \sqrt{3 - 4(2)(-3)}}{2(2)} = \frac{\sqrt{3} \pm \sqrt{3 + 24}}{4} = \frac{\sqrt{3} \pm \sqrt{27}}{4} = \frac{\sqrt{3} \pm 3\sqrt{3}}{4}$
$x_1 = \frac{4\sqrt{3}}{4} = \sqrt{3}$ (Not possible as $|\cos \theta| \le 1$)
$x_2 = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}$
So,$\cos \theta = -\frac{\sqrt{3}}{2}$.
The smallest positive angle $\theta$ is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

(B) Given the equation: $\cot \theta = \sin 2\theta$
Since $\cot \theta = \frac{\cos \theta}{\sin \theta}$ and $\sin 2\theta = 2 \sin \theta \cos \theta$,we have:
$\frac{\cos \theta}{\sin \theta} = 2 \sin \theta \cos \theta$
$\cos \theta = 2 \sin^2 \theta \cos \theta$
$\cos \theta (1 - 2 \sin^2 \theta) = 0$
This implies either $\cos \theta = 0$ or $1 - 2 \sin^2 \theta = 0$.
Case $1$: $\cos \theta = 0 \implies \theta = 90^o$ (within the principal range).
Case $2$: $2 \sin^2 \theta = 1 \implies \sin^2 \theta = \frac{1}{2} \implies \sin \theta = \pm \frac{1}{\sqrt{2}}$.
For $\sin \theta = \frac{1}{\sqrt{2}}$,$\theta = 45^o$.
Thus,the values are $45^o$ and $90^o$.

(A) Given equation: $\cos \theta + \sqrt{3} \sin \theta = 2$.
Divide the entire equation by $2$:
$\frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta = 1$.
Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we can rewrite this as:
$\sin \theta \cos \frac{\pi}{3} + \cos \theta \sin \frac{\pi}{3} = 1$.
This simplifies to:
$\sin(\theta + \frac{\pi}{3}) = 1$.
Since $\sin(\frac{\pi}{2}) = 1$,we have:
$\theta + \frac{\pi}{3} = \frac{\pi}{2}$.
$\theta = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$.
Wait,checking the options:
If $\theta = \frac{\pi}{3}$,then $\cos(\frac{\pi}{3}) + \sqrt{3} \sin(\frac{\pi}{3}) = \frac{1}{2} + \sqrt{3}(\frac{\sqrt{3}}{2}) = \frac{1}{2} + \frac{3}{2} = 2$.
Thus,$\theta = \frac{\pi}{3}$ is the correct solution.

(C) Given $\tan (\pi \cos \theta ) = \cot (\pi \sin \theta )$.
Using the identity $\cot(x) = \tan(\frac{\pi}{2} - x)$,we have $\tan (\pi \cos \theta ) = \tan \left( \frac{\pi}{2} - \pi \sin \theta \right)$.
This implies $\pi \cos \theta = n\pi + \frac{\pi}{2} - \pi \sin \theta$ for some integer $n$.
Dividing by $\pi$,we get $\cos \theta + \sin \theta = n + \frac{1}{2}$.
Since the range of $\sin \theta + \cos \theta$ is $[-\sqrt{2}, \sqrt{2}] \approx [-1.414, 1.414]$,the only possible value for $n$ is $0$,giving $\sin \theta + \cos \theta = \frac{1}{2}$.
We need to find $\sin \left( \theta + \frac{\pi}{4} \right) = \sin \theta \cos \frac{\pi}{4} + \cos \theta \sin \frac{\pi}{4}$.
$= \frac{1}{\sqrt{2}} (\sin \theta + \cos \theta) = \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{1}{2\sqrt{2}}$.

(A) Given,$\cot (\alpha + \beta ) = 0.$
Since $\cot \theta = 0$ implies $\theta = (2n + 1)\frac{\pi}{2}$ for $n \in I,$
we have $\alpha + \beta = (2n + 1)\frac{\pi}{2}.$
Now,consider $\sin (\alpha + 2\beta ) = \sin ((\alpha + \beta ) + \beta ).$
Substituting $\alpha + \beta = (2n + 1)\frac{\pi}{2},$
$\sin (\alpha + 2\beta ) = \sin ((2n + 1)\frac{\pi}{2} + \beta ) = \cos \beta$ if $n$ is even,or $-\cos \beta$ if $n$ is odd.
Alternatively,using the identity $\sin (\alpha + 2\beta ) = \sin (2(\alpha + \beta ) - \alpha ) = \sin ((2n + 1)\pi - \alpha ).$
Since $\sin ((2n + 1)\pi - \alpha ) = \sin (\pi - \alpha ) = \sin \alpha$ for even $n,$ and $\sin (3\pi - \alpha ) = \sin \alpha,$ the result is $\sin \alpha.$

(C) Given equation is $\cos x - \sin x = \frac{1}{\sqrt{2}}$.
Dividing both sides by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x = \frac{1}{2}$.
This can be written as $\cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4} = \frac{1}{2}$.
Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$,we get $\cos(x + \frac{\pi}{4}) = \frac{1}{2}$.
Since $\cos \frac{\pi}{3} = \frac{1}{2}$,the equation becomes $\cos(x + \frac{\pi}{4}) = \cos \frac{\pi}{3}$.
The general solution for $\cos \theta = \cos \alpha$ is $\theta = 2n\pi \pm \alpha$.
Therefore,$x + \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{3}$.
Case $1$: $x = 2n\pi + \frac{\pi}{3} - \frac{\pi}{4} = 2n\pi + \frac{\pi}{12}$.
Case $2$: $x = 2n\pi - \frac{\pi}{3} - \frac{\pi}{4} = 2n\pi - \frac{7\pi}{12}$.

(C) Given the equation: $12 \cot^2 \theta - 31 \csc \theta + 32 = 0$
Using the identity $\cot^2 \theta = \csc^2 \theta - 1$,we substitute:
$12(\csc^2 \theta - 1) - 31 \csc \theta + 32 = 0$
Simplify the equation:
$12 \csc^2 \theta - 12 - 31 \csc \theta + 32 = 0$
$12 \csc^2 \theta - 31 \csc \theta + 20 = 0$
Factor the quadratic equation in terms of $\csc \theta$:
$12 \csc^2 \theta - 16 \csc \theta - 15 \csc \theta + 20 = 0$
$4 \csc \theta (3 \csc \theta - 4) - 5 (3 \csc \theta - 4) = 0$
$(4 \csc \theta - 5)(3 \csc \theta - 4) = 0$
This gives two possible values for $\csc \theta$:
$\csc \theta = \frac{5}{4}$ or $\csc \theta = \frac{4}{3}$
Since $\sin \theta = \frac{1}{\csc \theta}$,we have:
$\sin \theta = \frac{4}{5}$ or $\sin \theta = \frac{3}{4}$.

(B) Given equation: $\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x$
Rearranging terms: $(\sin 3x + \sin x) - 3\sin 2x = (\cos 3x + \cos x) - 3\cos 2x$
Using sum-to-product formulas: $2\sin 2x \cos x - 3\sin 2x = 2\cos 2x \cos x - 3\cos 2x$
$\sin 2x(2\cos x - 3) = \cos 2x(2\cos x - 3)$
$(\sin 2x - \cos 2x)(2\cos x - 3) = 0$
Since $2\cos x - 3 \neq 0$ (as $\cos x$ cannot be $1.5$),we have $\sin 2x = \cos 2x$
$\tan 2x = 1$
$2x = n\pi + \frac{\pi}{4}$
$x = \frac{n\pi}{2} + \frac{\pi}{8}$

(B) Given equation: $5\cos^2 \theta + 7\sin^2 \theta - 6 = 0$
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we get:
$5(1 - \sin^2 \theta) + 7\sin^2 \theta - 6 = 0$
$5 - 5\sin^2 \theta + 7\sin^2 \theta - 6 = 0$
$2\sin^2 \theta - 1 = 0$
$2\sin^2 \theta = 1$
$\sin^2 \theta = \frac{1}{2}$
Since $\sin^2 \theta = \sin^2 \left(\frac{\pi}{4}\right)$,the general solution for $\sin^2 \theta = \sin^2 \alpha$ is $\theta = n\pi \pm \alpha$.
Therefore,$\theta = n\pi \pm \frac{\pi}{4}$.

(D) Divide the equation $a \cos x + b \sin x = c$ by $\sqrt{a^2 + b^2}$:
$\frac{a}{\sqrt{a^2 + b^2}} \cos x + \frac{b}{\sqrt{a^2 + b^2}} \sin x = \frac{c}{\sqrt{a^2 + b^2}}$
Let $\frac{a}{\sqrt{a^2 + b^2}} = \cos \alpha$ and $\frac{b}{\sqrt{a^2 + b^2}} = \sin \alpha$,where $\tan \alpha = \frac{b}{a}$.
Then the equation becomes $\cos(x - \alpha) = \frac{c}{\sqrt{a^2 + b^2}}$.
The general solution is $x - \alpha = 2n\pi \pm \cos^{-1} \left( \frac{c}{\sqrt{a^2 + b^2}} \right)$.
Substituting $\alpha = \tan^{-1} \left( \frac{b}{a} \right)$,we get:
$x = 2n\pi + \tan^{-1} \left( \frac{b}{a} \right) \pm \cos^{-1} \left( \frac{c}{\sqrt{a^2 + b^2}} \right)$.

(C) Given: $\sec 4\theta - \sec 2\theta = 2$
$\Rightarrow \frac{1}{\cos 4\theta} - \frac{1}{\cos 2\theta} = 2$
$\Rightarrow \frac{\cos 2\theta - \cos 4\theta}{\cos 4\theta \cos 2\theta} = 2$
$\Rightarrow \cos 2\theta - \cos 4\theta = 2 \cos 4\theta \cos 2\theta$
Using the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$\Rightarrow \cos 2\theta - \cos 4\theta = \cos 6\theta + \cos 2\theta$
$\Rightarrow - \cos 4\theta = \cos 6\theta$
$\Rightarrow \cos 6\theta + \cos 4\theta = 0$
Using the formula $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$\Rightarrow 2 \cos 5\theta \cos \theta = 0$
Case $1$: $\cos 5\theta = 0$ $\Rightarrow 5\theta = (2n + 1)\frac{\pi}{2}$ $\Rightarrow \theta = (2n + 1)\frac{\pi}{10}$
Case $2$: $\cos \theta = 0 \Rightarrow \theta = (2n + 1)\frac{\pi}{2} = n\pi + \frac{\pi}{2}$
Thus,the general solution is $\theta = n\pi + \frac{\pi}{2}$ or $\theta = \frac{n\pi}{5} + \frac{\pi}{10}$.

(A) Given equation: $\sin 2x + \sin 4x = 2\sin 3x$
Using the sum-to-product formula $\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$:
$2\sin(\frac{2x+4x}{2})\cos(\frac{2x-4x}{2}) = 2\sin 3x$
$2\sin 3x \cos(-x) = 2\sin 3x$
Since $\cos(-x) = \cos x$,we have:
$2\sin 3x \cos x - 2\sin 3x = 0$
$2\sin 3x(\cos x - 1) = 0$
This implies either $\sin 3x = 0$ or $\cos x = 1$.
If $\sin 3x = 0$,then $3x = n\pi \Rightarrow x = \frac{n\pi}{3}$.
If $\cos x = 1$,then $x = 2n\pi$.
Since $2n\pi$ is a subset of $\frac{n\pi}{3}$ (when $n$ is a multiple of $6$),the general solution is $x = \frac{n\pi}{3}$.

(B) Given the equation: $\tan (\cot x) = \cot (\tan x)$
Using the identity $\cot \theta = \tan (\frac{\pi}{2} - \theta)$,we have:
$\tan (\cot x) = \tan (\frac{\pi}{2} - \tan x)$
The general solution for $\tan \alpha = \tan \beta$ is $\alpha = n\pi + \beta$,where $n \in \mathbb{Z}$.
Thus,$\cot x = n\pi + \frac{\pi}{2} - \tan x$
Rearranging the terms:
$\cot x + \tan x = n\pi + \frac{\pi}{2}$
Using the identity $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\frac{1}{2} \sin 2x} = \frac{2}{\sin 2x}$:
$\frac{2}{\sin 2x} = n\pi + \frac{\pi}{2} = \frac{(2n + 1)\pi}{2}$
Solving for $\sin 2x$:
$\sin 2x = \frac{4}{(2n + 1)\pi}$

(B) Given $\cos p\theta = -\cos q\theta = \cos(\pi + q\theta)$.
This implies $p\theta = 2n\pi \pm (\pi + q\theta)$ for $n \in \mathbb{Z}$.
Case $1$: $p\theta = 2n\pi + \pi + q\theta \implies \theta = \frac{(2n + 1)\pi}{p - q}$. This forms an $A.P.$ with common difference $d_1 = \frac{2\pi}{|p - q|}$.
Case $2$: $p\theta = 2n\pi - (\pi + q\theta) \implies \theta = \frac{(2n - 1)\pi}{p + q}$. This forms an $A.P.$ with common difference $d_2 = \frac{2\pi}{p + q}$.
Comparing the two,the numerically smallest common difference is $\frac{2\pi}{p + q}$.

(D) The given equation is $a \sin x + b \cos x = c$.
Divide both sides by $\sqrt{a^2 + b^2}$:
$\frac{a}{\sqrt{a^2 + b^2}} \sin x + \frac{b}{\sqrt{a^2 + b^2}} \cos x = \frac{c}{\sqrt{a^2 + b^2}}$
Let $\frac{a}{\sqrt{a^2 + b^2}} = \cos \alpha$ and $\frac{b}{\sqrt{a^2 + b^2}} = \sin \alpha$.
Then the equation becomes $\sin(x + \alpha) = \frac{c}{\sqrt{a^2 + b^2}}$.
Since $|c| > \sqrt{a^2 + b^2}$,it follows that $\left| \frac{c}{\sqrt{a^2 + b^2}} \right| > 1$.
Because the range of the sine function is $[-1, 1]$,the equation $\sin(x + \alpha) = k$ has no solution if $|k| > 1$.
Therefore,the given equation has no solution.

(D) Given the equation $8\sec^2\theta - 6\sec\theta + 1 = 0$.
Let $x = \sec\theta$. The equation becomes $8x^2 - 6x + 1 = 0$.
Factoring the quadratic: $8x^2 - 4x - 2x + 1 = 0 \implies 4x(2x - 1) - 1(2x - 1) = 0$.
$(4x - 1)(2x - 1) = 0$.
So,$x = \frac{1}{4}$ or $x = \frac{1}{2}$.
This means $\sec\theta = \frac{1}{4}$ or $\sec\theta = \frac{1}{2}$.
Since the range of $\sec\theta$ is $(-\infty, -1] \cup [1, \infty)$,the values $\frac{1}{4}$ and $\frac{1}{2}$ are not possible.
Therefore,there are no real roots for $\theta$.
The number of roots is $0$.

(C) Given the equation: $3\sin^2x - 7\sin x + 2 = 0$
Factor the quadratic equation: $(3\sin x - 1)(\sin x - 2) = 0$
This gives two possibilities: $\sin x = \frac{1}{3}$ or $\sin x = 2$
Since the range of $\sin x$ is $[-1, 1]$,$\sin x = 2$ is impossible.
Therefore,we solve for $\sin x = \frac{1}{3}$.
In the interval $[0, 2\pi]$,there are $2$ solutions for $\sin x = \frac{1}{3}$.
In the interval $[0, 4\pi]$,there are $4$ solutions.
In the interval $[0, 5\pi]$,there are $4 + 2 = 6$ solutions,as the interval $[4\pi, 5\pi]$ contains $2$ additional solutions.

(A) Given equation: $\cos x + \cos 4x + \cos 2x + \cos 3x = 0$
Using the sum-to-product formula $\cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$:
$2 \cos \left(\frac{5x}{2}\right) \cos \left(\frac{3x}{2}\right) + 2 \cos \left(\frac{5x}{2}\right) \cos \left(\frac{x}{2}\right) = 0$
$2 \cos \left(\frac{5x}{2}\right) [\cos \left(\frac{3x}{2}\right) + \cos \left(\frac{x}{2}\right)] = 0$
Using $\cos C + \cos D = 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$:
$2 \cos \left(\frac{5x}{2}\right) [2 \cos x \cos \left(\frac{x}{2}\right)] = 0$
$4 \cos \left(\frac{5x}{2}\right) \cos x \cos \left(\frac{x}{2}\right) = 0$
Case $1$: $\cos \left(\frac{5x}{2}\right) = 0$ $\Rightarrow \frac{5x}{2} = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}$ $\Rightarrow x = \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5}$
Case $2$: $\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}$
Case $3$: $\cos \left(\frac{x}{2}\right) = 0$ $\Rightarrow \frac{x}{2} = \frac{\pi}{2}$ $\Rightarrow x = \pi$ (already included)
The distinct values are $\frac{\pi}{5}, \frac{\pi}{2}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{3\pi}{2}, \frac{9\pi}{5}$.
Total number of values is $7$.

(A) Given equation: $8 \cos x \left\{ \cos \left( \frac{\pi}{6} + x \right) \cos \left( \frac{\pi}{6} - x \right) - \frac{1}{2} \right\} = 1$
Using the identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$4 \cos x \left\{ \cos \left( \frac{\pi}{3} \right) + \cos(2x) - 1 \right\} = 1$
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$:
$4 \cos x \left\{ \frac{1}{2} + \cos 2x - 1 \right\} = 1$
$4 \cos x \left\{ \cos 2x - \frac{1}{2} \right\} = 1$
$4 \cos x \left\{ 2 \cos^2 x - 1 - \frac{1}{2} \right\} = 1$
$8 \cos^3 x - 6 \cos x = 1$
$2(4 \cos^3 x - 3 \cos x) = 1$
$2 \cos 3x = 1 \Rightarrow \cos 3x = \frac{1}{2}$
For $x \in [0, \pi]$,$3x \in [0, 3\pi]$.
Solutions for $3x$: $3x = \frac{\pi}{3}, 2\pi - \frac{\pi}{3}, 2\pi + \frac{\pi}{3} = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}$.
Thus,$x = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}$.
Sum of solutions $= \frac{\pi + 5\pi + 7\pi}{9} = \frac{13\pi}{9}$.
Given sum $= k\pi$,so $k = \frac{13}{9}$.