## Precalculus (6th Edition) Blitzer

The solutions to the given equation are $\frac{7\pi }{6},\,and\,\frac{11\pi }{6}$.
We have to solve the equation on the interval $[0,2\pi )$: $5\text{sin}\ x=2\text{co}{{\text{s}}^{2}}x-4$ And use the Pythagorean identity of trigonometry ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. And the expression ${{\cos }^{2}}x$ can be written as $1-{{\sin }^{2}}x$: \begin{align} & 5\text{sin}\ x=2\left( 1-\text{si}{{\text{n}}^{2}}x \right)-4 \\ & =2-2\text{si}{{\text{n}}^{2}}x-4 \\ & =-2{{\sin }^{2}}x-2 \end{align} After that, simplify it as: \begin{align} & 2{{\sin }^{2}}x+5\sin x+2=0 \\ & 2{{\sin }^{2}}x+4\sin x+\sin x+2=0 \end{align} And factorize the equation: \begin{align} & 2\sin x\left( \sin x+2 \right)+1\left( \sin x+2 \right)=0 \\ & \left( 2\sin x+1 \right)\left( \sin x+2 \right)=0 \end{align} And each factor needs to be calculated as: \begin{align} & 2\text{sin}\ x+1=0 \\ & 2\sin x=0-1 \\ & \text{sin}\ x=-\frac{1}{2} \end{align} Or, \begin{align} & \sin x+2=0 \\ & \sin x=0-2 \\ & =-2 \end{align} So, in the quadrant graph, the value of sine is $-\frac{1}{2}$ at $\frac{7\pi }{6}$, and $\frac{11\pi }{6}$. This implies, \begin{align} & sinx=sin\frac{7\pi }{6} \\ & x=\frac{7\pi }{6} \end{align} Then, take the value of $x$ as $\frac{11\pi }{6}$: \begin{align} & sinx=sin\frac{11\pi }{6} \\ & x=\frac{11\pi }{6} \end{align} And, So, in the quadrant graph, the value of sine is $-2$. This value of sine lies between the values of $-1$ and $1$. Therefore, the value of sine $-2$ will not be counted in the solution as it is less than the value of $-1$. Thus, the actual solution in the interval $[0,2\pi )$ will be $\frac{7\pi }{6},\,\text{ and }\,\frac{11\pi }{6}$.