Question

Solve the equation in the interval [0, 2pi]:
2sin^2x-5sinx+2=0

Answer 1

\[\Large
2 \sin ^2(x)-5 \sin
(x)+2=(\sin (x)-2) (2 \sin
(x)-1)=0\\
\Large
\sin(x)=\frac 1 2\\
\sin(x)=2\quad \text {( Impossible)}
\]

Answer 2

\[
\Large
\sin(x)=\frac 1 2\implies x=\frac \pi 6 + 2 k \pi \text{ or } x=\frac {5\pi}6 + 2 k \pi

\]
Choose the solutions in the given interval

Answer 3

For the actual solving, it may be easier to do it if you make a substitution first:
let \(u = \sin x\)
\[2\sin^2(x) - 5\sin(x) + 2 =0\]\[ 2u^2-5u+2=0\](solve by your favorite method)
\[u = 2, \,u=\frac{1}{2}\]Undo the substitution:
\[\sin(x) = 2\](impossible, as previously mentioned, so disregard)
\[\sin(x) = \frac{1}{2}\]\[x = \frac{\pi}{6},\,\frac{5\pi}{6}\]are the solutions in the specified interval.