Question

sin²(2x)=cos2x+4sin⁴x
помогите, пожалуйста

Answer 1

Решение во вложении...

Answer 2

$Sin^{2}2x=Cos2x+4Sin^{4}x\\\\Sin^{2} 2x-4Sin^{4}x=Cos2x\\\\4Sin^{2} xCos^{2}x-4Sin^{4}x=Cos2x\\\\4Sin^{2}x(Cos^{2}x-Sin^{2}x )-Cos2x=0\\\\4Sin^{2}x*Cos2x-Cos2x=0\\\\Cos2x(4Sin^{2}x-1)=0\\\\1)Cos2x=0\\\\2x=\frac{\pi }{2}+\pi n,nez\\\\x=\frac{\pi }{4}+\frac{\pi n }{2},nez\\\\2)4Sin^{2}x-1=0\\\\Sin^{2}x=\frac{1}{4}\\\\Sinx=\frac{1}{2} \\\\x_{1} =\frac{\pi }{6}+2\pi n,nez\\\\x_{2} =\frac{5\pi }{6}+2\pi n,nez\\\\Sinx=-\frac{1}{2}\\\\x_{3}=-\frac{\pi }{6} +2\pi n,nez\\\\x_{4}=-\frac{5\pi }{6}+2\pi$