Question

cos²($pi$-x)+sin2x=0

Answer 1

cos(π - x) = - cosx
(-cosx)^2 = cos^2(x)
sin(2x) = 2sinx*cosx
cos^2(x) + 2sinx*cosx = 0
cosx*(cosx + 2sinx) = 0
1) cosx = 0, x=π/2 + πk
2) -2sinx = cos
tgx = -1/2, x=arctg(-1/2) + πk
x = -arctg(1/2) + πk