Answer
The solution to the given equation is $\frac{\pi }{2}$.
Work Step by Step
We have to solve the equation on the interval $[0,2\pi )$:
$\begin{align}
& 2\text{si}{{\text{n}}^{2}}x=3-\text{sin}\ x \\
& 2\text{si}{{\text{n}}^{2}}x+\text{sin}\ x-3=0
\end{align}$
And simplify it as:
$\begin{align}
& 2{{\sin }^{2}}x-\sin x+3\sin x-3=0 \\
& 2\sin x\left( \sin x-1 \right)+3\left( \sin x-1 \right)=0 \\
& \left( \text{sin}x-1 \right)\left( 2\text{sin}x+3 \right)=0
\end{align}$
Each factor needs to be calculated as:
$\begin{align}
& \sin x-1=0 \\
& \sin x=0+1 \\
& \sin x=1
\end{align}$
Or
$\begin{align}
& 2\sin x+3=0 \\
& 2\sin x=0-3 \\
& \sin x=-\frac{3}{2}
\end{align}$
So, in the quadrant graph, the value of sine is $1$ at $\frac{\pi }{2}$. This implies,
$\begin{align}
& sinx=sin\frac{\pi }{2} \\
& x=\frac{\pi }{2}
\end{align}$
And in the quadrant graph, the value of sine counts to be $-\frac{3}{2}$. This value of sine lies between the values of $-1$ and $1$. Thus, the value of sine $-\frac{3}{2}$ 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 solution in the interval $[0,2\pi )$ will be $\frac{\pi }{2}$.
