3 sin^2 (2 п-2x) + 7 sin (п/2 - 2x) - 3=0

3 sin² (2 п-2x) + 7 sin (п/2 - 2x) - 3=0

3sin²2x+7cos2x-3=0

3 (1-cos²2x) + 7cos2x-3=0

3-3cos²2x+7cos2x-3=0|: (-1)

3cos²2x-7cos2x=0

t=cos2x

3t²-7t=0

t (3t-7) = 0

t=0 ∨ 3t-7=0

t=7/3

cos2x=0 cos2x≠7/3, т. к. |cos2x|≤1, а 7/3>1

2x=π/2+πn, n∈Z

x=π/4+πn/2, n∈Z