Simple Trig, got lost and need help.

Question

Solve the given equation over the interval [0,2): 2sinx+ (sqrt 3) sin^2x=-(sqrt 3) cos^2x

A. x=0, and x=(pi/2), and x=(3 pi/2), and x=2(pi)x=(3 pi/2), and x=(7 pi/2)

b. x=(4 pi/3) and x=(5 pi/3)

c. x=(7 pi/6) and x=(11 pi/6)

d. x=(pi/6) and x=(5 pi/6)

Answer 1

2sin x + √3 sin²x = -√3 cos²x

2sin x + √3 sin²x + √3 cos²x = 0

2sin x + √3 (sin²x + cos²x) = 0

2sin x + √3 = 0

 

Can you solve?