Answer
The solution is $\underline{x=\frac{\pi }{2},\frac{7\pi }{6},\frac{3\pi }{2},\frac{11\pi }{6}}$.
Work Step by Step
We have to solve the equation:
$\begin{align}
& \sin 2x+\cos x=0 \\
& 2\sin x\cos x+\cos x=0
\end{align}$
And use the formula,
$\sin 2\theta =2\sin \theta \cos \theta $
Factor:
$co\operatorname{s}x\left( 2\sin x+1 \right)=0$
From the above result it can be concluded that,
$\cos x=0$
Another term will be:
$2\sin x+1=0$
Compute the value of x:
$\begin{align}
& \cos x=0 \\
& x=\left( 2n+1 \right)\frac{\pi }{2}
\end{align}$
And the value of $\sin x$ will be:
$\begin{align}
& 2\sin x+1=0 \\
& 2\sin x=-1 \\
& \sin x=\frac{-1}{2}
\end{align}$
So, in the interval $\left[ 0,2\pi \right),$ the sine function is $\frac{-1}{2}$ at $\frac{7\pi }{6}\text{ and }\frac{11\pi }{6}$, according to the trigonometric table.
$\begin{align}
& \sin x=\frac{-1}{2} \\
& x=\frac{7\pi }{6}
\end{align}$
And another value is:
$\begin{align}
& \sin x=\frac{-1}{2} \\
& x=\frac{11\pi }{6}
\end{align}$
The period of the sine function is $2\pi $; the general solution of the equation is:
$x=\frac{7\pi }{6}+2n\pi $ and $x=\frac{11\pi }{6}+2n\pi $
To get different solutions, $\text{put }n=0,1,2,3\ldots $
$\begin{align}
& \text{For }n=0, \\
& x=\left( 2n+1 \right)\frac{\pi }{2} \\
& =\frac{\pi }{2}
\end{align}$
And the other value when n is 0:
$\begin{align}
& x=\frac{7\pi }{6}+2n\pi \\
& =\frac{7\pi }{6}
\end{align}$
Or
$\begin{align}
& x=\frac{11\pi }{6}+2n\pi \\
& =\frac{11\pi }{6}
\end{align}$
When the value of n is 1:
$\begin{align}
& \text{ }n=1 \\
& x=\left( 2n+1 \right)\frac{\pi }{2} \\
& =\frac{3\pi }{2}
\end{align}$
When we put other values of $n$, the solution becomes outside the interval $\left[ 0,2\pi \right).$
