Question

How do you solve `2sin^2x=Sinx` over the interval 0 to 2pi?

Answer 1

0, `pi`, `2pi`, `pi/6` and `5pi/6`,

Explanation:

Rearranging, sinx (2 sinx - 1) = 0.
In [0, 2`pi`],sin x = 0 for x = 0, `pi` and `2pi`.
The other factor = 0 gives the other two solutions.
sin `5pi/6` = sin (`pi - pi/6`) = sin `pi/6` = 1/2.