## Precalculus (6th Edition) Blitzer

The solutions to the given equation are $\frac{\pi }{6}\,\text{ and }\,\frac{5\pi }{6}$.
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 $\frac{\pi }{6}\,\text{ and }\,\frac{5\pi }{6}$.