Решение
Sin2x+4sin^2x=2cos^2x
4sin²x + 2sinxcosx - 2cos²x = 0
делим на cos²x ≠ 0
4tg²x + 2tgx - 2 = 0
делим на 2
2tg²x + tgx - 1 = 0
tgx = t
2t² + t - 1 = 0
D = 1 + 4*2*1 = 9
t₁ = (-1 - 3)/4
t₁ = - 1
t₂ = (-1 + 3)/4
t₂ = 1/2
1) tgx = - 1
x₁ = - π/4 + πk, k ∈ Z
2) tgx = 1/2
x₂ = arctg(1/2) + πn, n ∈ Z