Question

Find all solutions in the interval [0, 2π) for 2 sin^2 x = sin x

Answer 1

Answer:

D

Step-by-step explanation:

2 sin²x=sin x

2 sin² x-sin x=0

sin x(2 sin x-1)=0

sin x=0

x=0,π

2 sin x-1=0

2 sin x=1

sin x=1/2= sin π/6,sin (π-π/6)

x=π/6,5π/6