Question

помогите пожалуйста!
2sin(Pi+x)*sin(3Pi/2+x)+sinx=0

Answer 1

2sin(π+x)*sin(3π/2+x)+sinx=0
- 2sinx * (-cosx) + sinx = 0
sinx(2cosx + 1) = 0
1)  sinx = 0
x = πn, n∈Z
2)  2cosx + 1 = 0
cosx = -1/2
x = (+ -)*arccos(-1/2) + 2πk, k∈Z
x = (+ -)*(π - arccos(1/2)) + 2πk, k∈Z
x = (+ -)*(π - π/3) + 2πk, k∈Z
x = (+ -)*(2π/3) + 2πk, k∈Z