How do you solve $\cos x\left( 2\sin x+1 \right)=0?$

Hint: As in any trigonometric equations, there are three main steps for determining the trigonometric equation.
Step I: Find out the trigonometric value needed to solve the equation.
Step II: At last find the value of unknown that will result in the angle which we got in the second step. These are some important points or steps which we need to remember for solving the trigonometric equations.

Complete step by step solution:
Given equation:
$\cos \left( x \right)\left( 2\sin \left( x \right)+1 \right)=0$
Then,
Factor $\cos \left( x \right)$ out $2\cos \left( x \right)\sin \left( x \right)$
$\cos \left( x \right)\left[ 2\sin \left( x \right) \right]+\cos \left( x \right)=0$
Now, Raise $\cos \left( x \right)$ to prove of $1$
$\cos \left( x \right)\left[ 2\sin \left( x \right) \right]+\cos \left( x \right)=0$
Factor $\cos \left( x \right)$ out of ${{\cos }^{1}}\left( x \right)$
$\cos \left( x \right)\left( 2\sin \left( x \right) \right)+\cos \left( x \right)-1=0$
If any individual factor on left side of the equation is equal to $0.$ So, the whole expression will be equal to $0.$
$\cos \left( x \right)=0$
$2\sin \left( x \right)+1=0$
Set the factor which is equal to $0.$
$\cos \left( x \right)=0$
The exact value for $\left( 0 \right)$ is $\dfrac{\pi }{2}$
Then, the cosine function is positive in the first and quadrant fourth.
Now, subtract the reference angle from $2\pi $ to find a solution.
$x=2\pi -\dfrac{\pi }{2}$
Here we have to simplify
$x=2\pi -\dfrac{\pi }{2}$
To write $\dfrac{2\pi }{1}$ as in a fraction form with a common denominator and then multiply it by $\dfrac{2}{2}$
$x=\dfrac{2\pi }{1}\times \dfrac{2}{2}-\dfrac{\pi }{2}$
Now, Write each expression with a common denominator of $2,$ by multiplying each by an appropriate factors of $1.$
Combining it.
$\Rightarrow x=\dfrac{2\pi \times 2}{1\times 2}-\dfrac{\pi }{2}$
$\Rightarrow x=\dfrac{2\pi \times 2}{2}-\dfrac{\pi }{2}$
Also combine numerator over common denominators we get,
$x=\dfrac{2\pi \times 2-\pi }{2}$
Now, simplify it.
$x=\dfrac{4\pi -\pi }{2}$
$\Rightarrow x=\dfrac{3\pi }{2}$
For the period of function the formula is $\dfrac{2\pi }{\left| b \right|}$ and then replace $b$ as $1$ we get.
$\dfrac{2\pi }{\left| 1 \right|},$ Divide $\dfrac{2\pi }{2\pi }$ by $1$ we get,
Add $2\pi $ to every negative angle to get a positive angle.
$\dfrac{-\pi }{6}+2\pi $
$\therefore \dfrac{2\pi }{1}.\dfrac{6}{6}-\dfrac{\pi }{6}$ (Multiplied by $\dfrac{6}{6}$ then)
$\therefore \dfrac{2\pi -6}{1\times 16}-\dfrac{\pi }{6}$
$\Rightarrow \dfrac{2\pi .6}{6}-\dfrac{\pi }{6}=\dfrac{11\pi }{6}$
$\Rightarrow x=\dfrac{11\pi }{6}$

Finally we get the answer $\dfrac{\pi }{6},\dfrac{7\pi }{6},\dfrac{3\pi }{2}$ and $\dfrac{11\pi }{6}$

Additional Information:
Trigonometric equations are the equations which involve one or more unknown angles for the trigonometric ratios. Which is expressed as $\text{sine}\left( \sin \right),$ cosine $\left( \cos \right)$, tangent $\left( \tan \right)$, cotangent $\left( \cot \right),$ secant $\left( \sec \right)$, cosecant $\left( \text{cosec} \right)$ angles. There are various trigonometric equations formulas. Each of these trigonometric ratios are determined at different angles. If any equation involves the trigonometric function of a variable is known as trigonometric functions.
As for better understanding we take an.
Example as ${{\cos }^{2}}x+5\cos x-7=0$ or
$\sin 5x+3{{\sin }^{2}}x=6$ and many others.

Note: While using identities check the sign of what is in the given problem because there are various identities for various problems. The trigonometric involved in an equation. There are six categories of trigonometric identities. Each of these is a key trigonometric identity and we should memorize it. While verifying and proving any question we should use the identities which make two sides of a given equation identical in order to prove that it is true. There are some key points you should remember while solving any problem of trigonometry.