Trigonometrical Equations 1 Question 11

11. Let $S=x \in(-\pi, \pi): x \neq 0, \pm \frac{\pi}{2}$. The sum of all distinct solutions of the equation $\sqrt{3} \sec x+\operatorname{cosec} x$ $+2(\tan x-\cot x)=0$ in the set $S$ is equal to

(2016 Adv.)

(a) $-\frac{7 \pi}{9}$

(b) $-\frac{2 \pi}{9}$

(c) 0

(d) $\frac{5 \pi}{9}$

Show Answer

Answer:

Correct Answer: 11. (c)

Solution:

  1. Given, $\sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0$,

$(-\pi<x<\pi)-{0, \pm \pi / 2}$

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

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

Multiplying and dividing by $\sqrt{a^{2}+b^{2}}$, i.e. $\sqrt{3+1}=2$, we get

$ 2 \frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x \quad-2 \cos 2 x=0 $

$\Rightarrow \quad \cos x \cdot \cos \frac{\pi}{3}+\sin x \cdot \sin \frac{\pi}{3} \quad-\cos 2 x=0$

$ \begin{array}{ll} \Rightarrow & \cos \quad x-\frac{\pi}{3}=\cos 2 x \\ \therefore & 2 x=2 n \pi \pm x-\frac{\pi}{3} \quad \stackrel{\text { since, } \cos \theta=\cos \alpha}{\Rightarrow}=2 n \pi \pm \alpha \\ \Rightarrow & 2 x=2 n \pi+x-\frac{\pi}{3} \\ \text { or } & x=2 n \pi-x+\frac{\pi}{3} \\ \Rightarrow & x=2 n \pi-\frac{\pi}{3} \text { or } 3 x=2 n \pi+\frac{\pi}{3} \\ \Rightarrow & x=\frac{2 n \pi}{3}+\frac{\pi}{9} \end{array} $

$ \therefore \quad x=\frac{-\pi}{3} \text { or } x=\frac{\pi}{9}, \frac{-5 \pi}{9}, \frac{7 \pi}{9} $

Now, sum of all distinct solutions

$ =\frac{-\pi}{3}+\frac{\pi}{9}-\frac{5 \pi}{9}+\frac{7 \pi}{9}=0 $



NCERT Chapter Video Solution

Dual Pane