Question

Solve for $x:{\cos }^{4}x+{\sin }^{4}x=2\cos (2x+\frac{\pi }{6})\cos (2x-\frac{\pi }{6})$.

Answer 1

${\cos }^{4}x+{\sin }^{4}x=2\cos (2x+\frac{\pi }{6})\cos (2x-\frac{\pi }{6})$

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

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

$1-\frac{1-\cos 4x}{4}=\cos (\frac{\pi }{3})-\cos 4x$

$3+\cos 4x=2-4\cos 4x$

$5\cos 4x=-1$

$x={\cos }^{-1}\frac{1}{5}$