Question

Решить уравнение:
Sin^4x*cos^2х- cos^4x*sin^2x=cos2x

Answer 1

$sin^4x\cdot cos^2x-cos^4x\cdot sin^2x=cos2x\\\\sin^2x\cdot cos^2x\cdot (sin^2x-cos^2x)=cos2x\\\\(sinx\cdot cosx)^2\cdot (-cos2x)=cos2x\\\\-(\frac{1}{2}sin2x)^2\cdot cos2x=cos2x\\\\cos2x\cdot (1+\frac{1}{4}sin^22x)=0\\\\1)\; \; cos2x=0\; ,\; \; 2x=\frac{\pi}{2}+\pi n,\; n\in Z\\\\\underline {x=\frac{\pi}{4}+\frac{\pi n}{2}\; ,\; n\in Z}\\\\2)\; \; 1+\frac{1}{4}sin^22x=0\; ,\; \; sin^22x=-4\ \textless \ 0\; \; net\; reshenij\; ,\; t.k.\; sin^22x \geq 0$