Question

1) sin 2x + cos 2x= 0 2) sin 4x - cos 2x = 0

Answer 1

$\sin2x+\cos 2x=0|:\cos 2x \\ tg2x+1=0 \\ tg2x=-1 \\ 2x=- \frac{ \pi }{4} + \pi n, n \in Z \\ x=- \frac{ \pi }{8} + \frac{ \pi n}{2} , n \in Z$

$\sin 4x-\cos 2x=0 \\ 2\sin 2x\cos 2x-\cos 2x=0 \\ \cos 2x(2\sin2x-1)=0 \\ \left[\begin{array}{ccc}\cos 2x=0\\\sin 2x= \frac{1}{2} \end{array}\right\to \left[\begin{array}{ccc}2x= \frac{ \pi }{2}+ \pi n, n \in Z\\2x=(-1)^k\cdot \frac{ \pi }{3}+ \pi k, k \in Z \end{array}\right \\ \\ \left[\begin{array}{ccc}x_1= \frac{ \pi }{4} + \frac{ \pi n}{2} , n \in Z\\x_2=(-1)^k\cdot \frac{ \pi }{6} + \frac{ \pi k}{2},k \in Z\end{array}\right$