QJEEAdv16Ma0ad2: JEE Advanced 2016 Mathematics - Class 11 JEEAdv16Ma0ad2 Solution: Let The sum of all distinct solutions of the equation in the…

Question

Verified Solution • 2025–26 Session moderate

Let

S = \left\{ {x \in \left( { - \pi ,\pi } \right):x \ne 0, \pm {\pi \over 2}} \right\}.

The sum of all distinct solutions of the equation

\sqrt 3 \,\sec x + \cos ec\,x + 2\left( {\tan x - \cot x} \right) = 0

in the set S is equal to

Accepted Answer

Let us consider

S=\left\{x \in(-\pi, \pi), x \neq 0, \pm \frac{\pi}{2}\right\}

The given equation is

\begin{aligned}&amp; \sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0 \\\\&amp; \Rightarrow \frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}+2\left(\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}\right)=0 \\\\&amp; \Rightarrow \sqrt{3} \sin x+\cos x+2\left(\sin ^2 x-\cos ^2 x\right)=0 \\\\&amp; \Rightarrow \sqrt{3} \sin x+\cos x=2 \cos 2 x \\\\&amp; \Rightarrow \frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=\cos 2 x \\\\&amp; \Rightarrow \cos x \cos \frac{\pi}{3}+\sin x \sin \frac{\pi}{3}=\cos 2 x \\\\&amp; \Rightarrow \cos 2 x=\cos \left(x-\frac{\pi}{3}\right) \\\\&amp; \Rightarrow 2 x=2 n \pi \pm\left(x-\frac{\pi}{3}\right) \quad\quad(x \in I)\end{aligned}

Case 1: When

2 x=2 n \pi+x-\frac{\pi}{3},

we have

x=2 n \pi-\frac{\pi}{3}

.

If

n=0

, we get

x=-\frac{\pi}{3}

.

If

n=1

, we get

x=2 \pi-\frac{\pi}{3}

.

If

n=-1, x=-2 \pi-\frac{\pi}{3}

.

Case 2: When

2 x=2 n \pi-x+\frac{\pi}{3}

, we get

x=\frac{2 n \pi}{3}+\frac{\pi}{9}

.

If

n=0

, we get

x=\frac{\pi}{9}

.

If

n=1

, we get

x=\frac{2 \pi}{3}+\frac{\pi}{9}

.

If

n=2

, we get

n=2 x=\frac{4 \pi}{3}+\frac{\pi}{9}

.

If

n=-1

, we get

x=\frac{-2 \pi}{3}+\frac{\pi}{9}

.

Therefore, the sum of all distinct solutions of the given equation is

\frac{-\pi}{3}+\frac{\pi}{9}+\frac{2 \pi}{3}+\frac{\pi}{9}-\frac{2 \pi}{3}+\frac{\pi}{9}=0

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