Discussion Forum

1)

If sum of all the solutions of the equation

$8\cos x.( \cos ( \frac{\pi}{6}+x).\cos ( \frac{\pi}{6}-x)-\frac{1}{2})=1$ in [0,π ] is kπ , then k is equal to

Answer:

Option B

Explanation:

Key Idea Apply the  identity 

$\cos ( x+y)\cos ( x-y)=\cos^{2}x-\sin^{2}y$ and   $\cos 3x=4\cos^{3}x-3\cos x$

We have,

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

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

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

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

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

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

$\Rightarrow$ $2\cos3x=1\Rightarrow\cos 3x=\frac{1}{2}$

$\Rightarrow$ 

$$3x=\frac{\pi}{3},\frac{5\pi}{3},\frac{7\pi}{3}$$    [0≤ 3x ≤ 3π ]

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

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

Hence    k= $\frac{13}{9}$