Question

The number of solutions of the equation $\cos \left(x+\frac{\pi }{3}\right)\cos \left(\frac{\pi }{3}-x\right)=\frac{1}{4}{\cos }^{2}2x,x\in [-3\pi ,3\pi ]$ is :

• A. 8
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (D)

$\cos \left(\frac{\pi }{3}+x\right)\cos \left(\frac{\pi }{3}-x\right)=\frac{1}{4}{\cos }^{2}2x$
$x\in [-3\pi ,3\pi ]$
$4\left({\cos }^2\left(\frac{\pi }{3}\right)-{\sin }^{2}x\right)={\cos }^{2}2x$
$4\left(\frac{1}{4}-{\sin }^{2}x\right)={\cos }^{2}2x$
$1-4{\sin }^{2}x={\cos }^{2}2x$
$1-2(1-\cos 2x)={\cos }^{2}2x$
let $\cos 2x=t$
$-1+2\cos 2x={\cos }^{2}2x$
$t^2-2t+1=0$
$(t-1{)}^2=0$
$t=1\cos 2x=1$
$2x=2n\pi$
$x=n\pi$
$n=-3,-2,-1,0,1,2,3$