Question

The values of '$\theta$' satisfying the equation $\cos \theta .\cos \left(\frac{\pi }{3}+\theta \right).\cos \left(\frac{\pi }{3}-\theta \right)=\frac{1}{4}$ is

Answer 1

As we know that $\cos (A+B)\cos (A-B)={\cos }^{2}B-{\sin }^{2}A$

Given $\cos (\frac{\pi }{3}+\theta )\cos (\frac{\pi }{3}-\theta )=\frac{1}{4}$

$⟹{\cos }^2\theta -{\sin }^2\frac{\pi }{3}=\frac{1}{4}$

$⟹{\cos }^2\theta =\frac{1}{4}+\frac{3}{4}$

$⟹{\cos }^2\theta =1$

$⟹\theta =n\pi$