Question

$3 sin^{2} 2x+7cos2x-3=0$

Answer 1

3sin²2x + 7cos2x - 3=0
3(1 - cos²2x) + 7cos2x - 3=0
3 - 3cos²2x + 7cos2x -3=0
-3cos²2x + 7cos2x=0
-cos2x(3cos2x - 7)=0
cos2x = 0        3cos2x-7=0
2x=π/2 + πn        cos2x=7/3 >1 ОДЗ
x = π/4 + πn/2, n∈Z
Ответ: π/4 + πn/2, n∈Z