Sum of Distinct Solutions of Trigonometric Equation

To find the sum of all distinct solutions of the given trigonometric equation in the set $S = \left\{ x \in (-\pi, \pi) : x \neq 0, \pm \frac{\pi}{2} \right\}$, we begin by expressing all trigonometric functions in terms of $\sin x$ and $\cos x$:

$\sqrt{3} \sec x + \csc x + 2(\tan x - \cot x) = 0$

Substituting $\sec x = \frac{1}{\cos x}$, $\csc x = \frac{1}{\sin x}$, $\tan x = \frac{\sin x}{\cos x}$, and $\cot x = \frac{\cos x}{\sin x}$:

$\frac{\sqrt{3}}{\cos x} + \frac{1}{\sin x} + 2\left(\frac{\sin x}{\cos x} - \frac{\cos x}{\sin x}\right) = 0$

Combining the fractions over the common denominator $\sin x \cos x$:

$\frac{\sqrt{3}\sin x + \cos x + 2(\sin^2 x - \cos^2 x)}{\sin x \cos x} = 0$

Since $x \in S$, we have $\sin x \neq 0$ and $\cos x \neq 0$ (i.e., $x \neq 0, \pm \frac{\pi}{2}$). Thus, we set the numerator to zero:

$\sqrt{3}\sin x + \cos x + 2(\sin^2 x - \cos^2 x) = 0$

Using the double-angle identity $\cos 2x = \cos^2 x - \sin^2 x$, the equation becomes:

$\sqrt{3}\sin x + \cos x - 2\cos 2x = 0$

Dividing the entire equation by $2$:

$\frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x = \cos 2x$

Rewriting the left side using the cosine addition formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$ with $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$:

$\cos\left(x - \frac{\pi}{3}\right) = \cos 2x$

The general solution for $\cos \theta = \cos \alpha$ is $\theta = 2k\pi \pm \alpha$ for $k \in \mathbb{Z}$. Therefore:

$2x = 2k\pi \pm \left(x - \frac{\pi}{3}\right), \quad k \in \mathbb{Z}$

We split this into two cases:

Case 1: Taking the positive sign: $2x = 2k\pi + x - \frac{\pi}{3} \implies x = 2k\pi - \frac{\pi}{3}$

For $x \in (-\pi, \pi)$, the only integer value $k = 0$ yields: $x = -\frac{\pi}{3}$

Case 2: Taking the negative sign: $2x = 2k\pi - \left(x - \frac{\pi}{3}\right) \implies 3x = 2k\pi + \frac{\pi}{3} \implies x = \frac{(6k + 1)\pi}{9}$

For $x \in (-\pi, \pi)$, we test integer values of $k$:

• For $k = 0$: $x = \frac{\pi}{9} \in S$
• For $k = 1$: $x = \frac{7\pi}{9} \in S$
• For $k = -1$: $x = -\frac{5\pi}{9} \in S$

For all other values of $k$, $x$ falls outside the interval $(-\pi, \pi)$.

Thus, the set of all distinct solutions in $S$ is: $\left\{ -\frac{3\pi}{9}, -\frac{5\pi}{9}, \frac{\pi}{9}, \frac{7\pi}{9} \right\}$

Summing these solutions together: $\text{Sum} = \left(-\frac{3\pi}{9}\right) + \left(-\frac{5\pi}{9}\right) + \frac{\pi}{9} + \frac{7\pi}{9} = \frac{-3 - 5 + 1 + 7}{9}\pi = 0$

Hence, the correct option is C.