Trigonometrical Equations 1 Question 8

8. If sum of all the solutions of the equation

$ 8 \cos x \cdot \cos \frac{\pi}{6}+x \cdot \cos \frac{\pi}{6}-x-\frac{1}{2}=1 $

in $[0$,

(a) $\frac{2}{3}$

(b) $\frac{13}{9}$

(c) $\frac{8}{9}$

(d) $\frac{20}{9}$

(2018 Main)

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Answer:

Correct Answer: 8. (b)

Solution:

  1. Key idea Apply the identity

$\cos (x+y) \cos (x-y)=\cos ^{2} x-\sin ^{2} y$

and $\cos 3 x=4 \cos ^{3} x-3 \cos x$

We have, $8 \cos x \cos \frac{\pi}{6}+x \cos \frac{\pi}{6}-x-\frac{1}{2}=1$

$\Rightarrow 8 \cos x \cos ^{2} \frac{\pi}{6}-\sin ^{2} x-\frac{1}{2}=1$

$\Rightarrow \quad 8 \cos x \frac{3}{4}-\sin ^{2} x-\frac{1}{2}=1$

$\Rightarrow 8 \cos x \frac{3}{4}-\frac{1}{2}-1+\cos ^{2} x=1$

$\Rightarrow \quad 8 \cos x \frac{-3+4 \cos ^{2} x}{4}=1$

$\Rightarrow \quad 2\left(4 \cos ^{3} x-3 \cos x\right)=1$

$\Rightarrow \quad 2 \cos 3 x=1 \Rightarrow \cos 3 x=\frac{1}{2}$

$\Rightarrow \quad 3 x=\frac{\pi}{3}, \frac{5 \pi}{3}, \frac{7 \pi}{3} \quad[0 \leq 3 x \leq 3 \pi]$

$\Rightarrow \quad x=\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}$

Sum $=\frac{\pi}{9}+\frac{5 \pi}{9}+\frac{7 \pi}{9}=\frac{13 \pi}{9} \Rightarrow k \pi=\frac{13 \pi}{9}$

Hence, $\quad k=\frac{13}{9}$



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