Question

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

Answer 1

Решение во вложении...

Answer 2

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