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.