Answer
The solutions to the given equation are $\frac{7\pi }{6},\,and\,\frac{11\pi }{6}$.
Work Step by Step
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 $\frac{7\pi }{6},\,\text{ and }\,\frac{11\pi }{6}$.
