Решить уравнение sin^2x+cos^22x=1

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