Решить уравнение sin^2x+cos^22x=1

sin²x + cos²2x = 1 -1 + cos²2x + sin²x = 0 -sin²2x + sin²x = 0 -sin2x•sin2x + sin²x = 0 -4sin²x•cos²x + sin²x = 0 sin²x(-4cos²x + 1) = 0 sin²x = 0 и -4cos²x = -1 sinx = 0 и cos²x = 1/4 sinx = 0 x = πn, n ∈ Z cosx = 1/2 и cosx = -1/2 x = ±π/3 + 2πn, n ∈ Z и x = ±2π/3 + 2πn, n ∈ Z.