Answer
The solutions to the given equation are $\frac{\pi }{6}\,\text{ and }\,\frac{5\pi }{6}$.
Work Step by Step
We solve the equation on the interval $[0,2\pi )$:
$\begin{align}
& 2\text{si}{{\text{n}}^{2}}x=2-3\text{sin}x \\
& 2\text{si}{{\text{n}}^{2}}x+3\text{sin}x-2=0
\end{align}$
And simplify it as:
$\begin{align}
& 2{{\sin }^{2}}x+4\sin x-\sin x-2=0 \\
& 2\sin x\left( \sin x+2 \right)-1\left( \sin x+2 \right)=0 \\
& \left( 2\text{sin}x-1 \right)\left( \text{sin}x+2 \right)=0
\end{align}$
Each factor needs to be calculated as:
$\begin{align}
& 2\sin x-1=0 \\
& 2\sin x=0+1 \\
& \sin x=\frac{1}{2}
\end{align}$
Or
$\begin{align}
& \sin x+2=0 \\
& \sin x=0-2 \\
& \sin x=-2
\end{align}$
In the quadrant graph, the value of sine is $\frac{1}{2}$ at $\frac{\pi }{6}$ and $\frac{5\pi }{6}$. This implies,
$\begin{align}
& sinx=sin\frac{\pi }{6} \\
& x=\frac{\pi }{6}
\end{align}$
$\begin{align}
& sinx=sin\frac{5\pi }{6} \\
& x=\frac{5\pi }{6}
\end{align}$
And in the quadrant graph, the value of sine is $-2$. This value of sine does not lie between the values of $-1$ and $1$. Thus, the value of sine $-2$ will not be counted in the solution as it is less than the value of $-1$.
These are the proposed solutions of the sine function. Thus, the actual solutions in the interval $[0,2\pi )$ will be $\frac{\pi }{6}\,\text{ and }\,\frac{5\pi }{6}$.
