Question

Помогите sin2x=cos^2x

Answer 1

$\sin2x=\cos^2x\\2\sin x\cos x-\cos^2x=0\\\cos x(2\sin x-\cos x)=0\\\cos x=0\Rightarrow x=\frac\pi2+\pi n\\\\2\sin x-\cos x=0\\2\sin x=\cos x\\2tgx=1\\tgx=\frac12\\x=arctg\frac12+\pi n,\;n\in\mathbb{Z}$