Animated Solution for Mathematics - Trigonometry: Let $S=\{x\in (-\pi ,\pi ):x\ne 0,\pm \pi /2\}$. The sum of all distinct solutions of the equation $\sqrt{3}\sec x+\csc x+2(\tan x-\cot x)=0$ in the set $S$ is equal to

Imagine you are standing at the edge of a dense, tangled forest. This forest is the equation:

At first glance, it looks chaotic—a mix of secants, cosecants, tangents, and cotangents. Many students panic here, trying to apply identities blindly. But as an elite aspirant, you know better; you don't fight the chaos, you simplify it.

Respecting the Boundaries

Before we touch a single variable, we must look at the domain $S = \{ x \in (-\pi, \pi) : x eq 0, \pm \frac{\pi}{2} \}$.

Think of the unit circle. At $x=0$, $\csc x$ and $\cot x$ are undefined. At $x=\pm \frac{\pi }{2}$, $\sec x$ and $\tan x$ are undefined.

These are the "forbidden zones." If our algebra leads us to these values, we must reject them immediately as they are mathematical mirages.

The Great Conversion

Now, let's strip away the disguises. We know that $\sec x=\frac{1}{\cos x}$, $\csc x=\frac{1}{\sin x}$, $\tan x=\frac{\sin x}{\cos x}$, and $\cot x=\frac{\cos x}{\sin x}$.

Substituting these into our equation gives us:

It looks messier, but this is a strategic mess. We have moved from four different functions to just two: sine and cosine, which is the common language of trigonometry.

The Algebraic Dance

To clean this up, we need a common denominator. The least common multiple is $\sin x\cos x$. Multiplying through, we get:

For a fraction to be zero, the numerator must be zero. We can safely ignore the denominator, provided we remember our domain constraints.

Now, look at the term $2({\sin }^{2}x-{\cos }^{2}x)$. Recall the double angle identity: $\cos 2x={\cos }^{2}x-{\sin }^{2}x$. Therefore, ${\sin }^{2}x-{\cos }^{2}x=-\cos 2x$.

Our equation transforms into:

The Identity Revelation

We are almost there. Rearranging gives us $\sqrt{3}\sin x+\cos x=2\cos 2x$. This is a classic form: $a\sin x+b\cos x$.

To solve this, we divide by $\sqrt{a^2+b^2}$, which is $\sqrt{(\sqrt{3}{)}^2+1^2}=2$. Dividing the entire equation by $2$, we get:

Recognize the values: $\frac{\sqrt{3}}{2}=\sin \frac{\pi }{3}$ and $\frac{1}{2}=\cos \frac{\pi }{3}$. The left side becomes $\sin \frac{\pi }{3}\sin x+\cos \frac{\pi }{3}\cos x$, which is exactly $\cos (x-\frac{\pi }{3})$.

We have arrived at the elegant core:

The Final Sum

We use the general solution for $\cos \theta =\cos \alpha$, which is $\theta =2n\pi \pm \alpha$. This gives us two cases:

Solving these for $n=0,1,-1$ yields our distinct solutions: $-\frac{\pi }{3}$, $\frac{\pi }{9}$, $\frac{7\pi }{9}$, and $-\frac{5\pi }{9}$.

When we sum these, the symmetry takes over:

The complexity collapses into 0. This is the beauty of mathematics—the most intricate problems often resolve into the most elegant answers.