Find all solutions in the interval [0, 2π).

2 sin2x = sin x

Answers:

x = π/3, 2π/3

x = π/2, 3π/2, π/3, 2π/3

x = 0, π, π/6, 5π/6

x = π/6, 5π/6

2sin^2 (x) = sinx

we switch sinx from left to right and take sinx in front of brackets.

sinx * (2sinx - 1) = 0

now we need to solve 2 cases.

first:

sinx = 0

that is for x = 0 and x = pi

second:

2sinx - 1 = 0

2sinx = 1

sinx = 1/2

x = pi/6 or 5pi/6

the case we had here is something like a*b=0 so either a=0 or b=0 thats why we solved problem this way.

At the end comparing our results with options we can see that answer is third option.