Question

Sin2x+cos4x= помогите решитт пожалуйста

Answer 1

$sin2x+cos4x=0\\sin2x+cos^22x-sin^22x=0\\sin2x+1-2sin^22x=0\\2sin^22x-sin2x-1=0;sin2x=a;-1\ \textless \ a\ \textless \ 1\\2a^2-a-1=0\\a_{1,2}=\frac{1^+_-3}{4}\\a_1=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_2=-\frac{1}{2}\\sin2x=1\ \ \ \ \ \ \ \ \ \ \ sin2x=-\frac{1}{2}\\2x=\frac{\pi}{2}+2\pi n\ \ \ \ \ \ 2x=-\frac{\pi}{6}+2\pi n\ \ \ \ \ \ \ \ \ 2x=\frac{7\pi}{6}+2\pi n;n\in Z\\x=\frac{\pi}{4}+\pi n\ \ \ \ \ \ \ \ \ x=-\frac{\pi}{12}+\pi n\ \ \ \ \ \ \ \ \ \ \ x=\frac{7\pi}{12}+\pi n;n\in Z$