Question

8 sin^2 5x+ 6 cos 5x-3=0 помогите пожалуйста...

Answer 1

sin^2(5x) = 1 - cos^2(5x)
8 - 8cos^2(5x) + 6cos5x - 3 = 0
8cos^2(5x) - 6cos5x -5 = 0
t = cos5x
8t^2 - 6t - 5 = 0
t = (6 +-√(36 + 160)) / 16 = (6 +- 14) / 16
t = 20/16 не подходит, так как cos5x <= 1
t = -8/16 = -1/2
cos5x = -1/2
5x = +-2π/3 + 2πn, n ∈ Z
x = +- 2π/15 + 2πn / 5, n ∈ Z