Sin^4x-cos^4x=sin2x-1/2

Решите задачу:

$sin^4x-cos^4x=sin2x-\frac{1}{2}\\\\\underbrace {(sin^2x-cos^2x)}_{-cos2x}\underbrace {(sin^2x+cos^2x)}_{1}=sin2x-\frac{1}{2}\\\\-cos2x=sin2x-\frac{1}{2}\\\\sin2x+cos2x=\frac{1}{2}\\\\\sqrt2\cdot (\frac{1}{\sqrt2}sin2x+\frac{1}{\sqrt2}cos2x)=\frac{1}{2}\; ,\; \; \; \; \frac{1}{\sqrt2}=\frac{\sqrt2}{2} \\\\\sqrt2\cdot (cos\frac{\pi}{4}\cdot sin2x+sin\frac{\pi}{4}\cdot cos2x)=\frac{1}{2}\\\\\sqrt2\cdot sin(2x+\frac{\pi}{4})=\frac{1}{2}\\\\sin(2x+\frac{\pi}{4})=\frac{1}{2\sqrt2}$

$2x+\frac{\pi}{4}=(-1)^{n}\cdot arcsin\frac{1}{2\sqrt2}+\pi n,\; n\in Z\\\\2x=-\frac{\pi}{4}+(-1)^{n}\cdot arcsin\frac{1}{2\sqrt2}+\pi n,\; n\in Z\\\\x=-\frac{\pi}{8}+\frac{(-1)^{n}}{2}\cdot arcsin\frac{1}{2\sqrt2}+\frac{\pi n}{2},\; n\in Z$