Question

3 sin x/5 =2 cos^2 (x/5+П) помогите решить пожалуйста...

Answer 1

cos(π + x/5) = -cos(x/5)
(-cos(x/5))^2 = cos^2(x/5)
cos^2(x/5) = 1 - sin^2(x/5)
3sin(x/5) = 2 - 2sin^2(x/5)
2sin^2(x/5) + 3sin(x/5) - 2 = 0
t = sin(x/5)
2t^2 + 3t - 2 = 0
t = (-3 +- √(9 + 16)) / 4 = (-3 +- 5) / 4
t = -2 не подходит, так как cos(x/5) >= -1
t = 2/4 = 1/2
cos(x/5) = 1/2
x/5 = +-π/3 + 2πn, n ∈ Z
x = +- 5π/3 + 10πn, n ∈ Z