Решите уравнение 2sin^2x-cos2x=1

2sin^2x - cos2x = 1

2sin^2x - (1 - 2sin^2x) - 1 = 0

2sin^2x - 1 + 2sin^2x - 1 = 0

4sin^2x - 2 = 0

4sin^2x = 2

sin^2x = 1/2

sinx = ± √2/2

1) sinx = √2/2

x = pi/4 + 2pik, k ∈Z

x = 3pi/4 + 2pik, k ∈Z

2) sinx = - √2/2

x = - pi/4 + 2pik, k ∈ Z

x = 5pi/4 + 2pik, k ∈Z