A Case With No Angle

$2{\sin\left(x\right)}^{2}-5\sin\left(x\right)+2 = 0$ factors as $\left(2\sin\left(x\right)-1\right)\left(\sin\left(x\right)-2\right) = 0$. The first case gives sin x = 1/2: 30° and 150°. The second gives sin x = 2.

No angle has sine 2: a sine is the height of a point on a circle of radius 1, so it stays from −1 to 1. That case drops out, and the answers are 30° and 150°. Check each case's value before reaching for arcsin.