Question

Solve on the interval [0, 2pi):

2sin^2 x-3sin x +1=0

Please explain answer. Thank you!

Answer 1

Factor it.

Answer 2

If you let \(u=\sin x\) then you can more easily see how it is factored: \[
2u^2-3u+1 = (2u-1)(u-1)
\]Which becomes: \[
(2\sin x -1)(\sin x -1)=0
\]

Answer 3

Each factor is set to \(0\), and then you solve.