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.