JEE Main 2018 — Trigonometric Equations Question with Solution

Given equation $8\cos x.\left(\cos \left(\frac{\pi }{6}+x\right).\cos \left(\frac{\pi }{6}-x\right)-\frac{1}{2}\right)=1$

We would use the formula

$\cos \left(A+B\right).\cos \left(A-B\right)={\cos }^{2}A-{\sin }^{2}B$

Now, $\cos \left(\frac{\pi }{6}+x\right).\cos \left(\frac{\pi }{6}-x\right)={\cos }^2\frac{\pi }{6}-{\sin }^{2}x$

$=\frac{3}{4}-{\sin }^{2}x$

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

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

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

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

$\Rightarrow 8{\cos }^{3}x-6\cos x=1$

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

$\Rightarrow 2\cos 3x=1$

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

General Solution $3x=2n\pi \pm \frac{\pi }{3}$

$\Rightarrow x=\frac{2n\pi }{3}\pm \frac{\pi }{9}$

$x\in \left[0, \pi \right]$

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

$\mathrm{Sum}=\frac{13\pi }{9}, k=\frac{13}{9}$