Question

Cos^2(x)+4sin^2(x)=2sin(2x)

Answer 1

$\cos^2x+4\sin^2x=2\sin2x\\ \\ \cos^2x+4\sin^2x=4\sin x\cos x|:\cos^2x\\ \\ 1+4tg^2x=4tgx\\ \\ 4tg^2x-4tg x+1=0\\ \\(2tgx-1)^2=0\\ \\ 2tg x-1=0\\ \\ tgx=0.5\\ \\ x=arctg0.5+ \pi n,n \in Z$