Sin^4x+cos^4x=cos^2 2 x

Решите задачу:

$sin^4x+cos^4x=cos^22x\\\\sin^4x+cos^4x=(cos^2x-sin^2x)^2\\\\\star [\, cos2x=cos^2x-sin^2x\, ]\star \\\\sin^4x+cos^4x=cos^4x-2sin^2x\cdot cos^2x+sin^4x\\\\0=-2sin^2x\cdot cos^2x\\\\sin^2x\cdot cos^2x=0\\\\\star [\, 2sinx\cdot cosx=sin2x\, ,\; 4sin^2x\cdot cos^2x=sin^22x]\star \\\\\frac{sin^22x}{4}=0\\\\sin^22x=0\; \; \Rightarrow sin2x=0\\\\2x=\pi n ,\; n\in Z\\\\x=\frac{\pi n}{2}\; ,\; n\in Z$