Find all real numbers in the interval 02pi that satisfy the

Find all real numbers in the interval [0,2pi] that satisfy the equation

2 sin²x = sin x

Please show how to solve the answer is (0, pi, pi/6, 5pi/6)

Solution

2Sin2X = SinX

2Sin2X - SinX = 0

By substituting the formula for Sin2X = 2SinXCosX, we get
4SinXCosX - SinX = 0
Taking SinX common, we get
SinX (4CosX - 1) = 0
i.e;

Either SinX = 0 or Cos X = 1/4

Taking SinX = 0; We get X = 0 or Pi

Taking CosX = 1/4, We get X = 5Pi/12 since Cos^-1 (1/4) = 75⁰

So X = 0, Pi, 5Pi/12