Trigonometrical Equations Ques 30
- 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: 30.(b)
Solution:
- 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, $\quad k=\frac{13}{9}$
BETA
