Question

Cos^2*2x-Sin^2*2x=0
Помогите пожалуйста

Answer 1

$cos^22x-sin^22x=0\\\\cos4x=0\\\\4x=\frac{\pi}{2}+\pi n,\; n\in Z\\\\x=\frac{\pi}{8}+\frac{\pi n}{4}\; ,\; n\in Z\\\\\\\star \; \; cos2\alpha =cos^2\alpha -sin^2\alpha \; \; \star \; \; \; \; \; \; \; \; \; \alpha =2x\; \; \to \; \; 2\alpha =4x$