Solve \[2{\cos ^2}x + 3\sin x = 0\]

Hint:
To solve the above equation, firstly we will make it equation of only $\sin x$ or $\cos x$ function with the help of ${\sin ^2}x + {\cos ^2}x = 1$ identity then we will solve this equation as a quadratic equation by finding its factors.

Complete step by step solution:
\[2{\cos ^2}x + 3\sin x = 0......(1)\]
We know that, in trigonometry we have and identity${\sin ^2}x + {\cos ^2}x = 1$, which we can mold as per our requirement as ${\cos ^2}x = 1 - {\sin ^2}x$and put in the equation (1) in the place of ${\cos ^2}x$
$2(1 - {\sin ^2}x) + 3\sin x = 0$
Now we will simplify the above equation by opening the bracket.
$2 - 2{\sin ^2}x + 3\sin x = 0$
We will keep $\sin x$terms on left side and constant on right side
$2{\sin ^2}x - 3\sin x = 2$
Now we will put $\sin x = y$to make it quadratic equation
$2{y^2} - 3y - 2 = 0$
To solve quadratic equation, we need to make factors, which on multiplication give -4 and on subtraction give -3
$2{y^2} - 4y + y - 2 = 0$
Take $ + 2y$common from first two terms and +1 common from last two terms
$2y(y - 2) + 1(y - 2) = 0$
Now, we will take $(y - 2)$common from above equation
$(y - 2)(2y + 1) = 0$
From above equation, we get two solutions which are as follows:
$y = 2$and $y = - \dfrac{1}{2}$
Here, we will put the real of $y = \sin x$
$\sin x = 2$and $\sin x = - \dfrac{1}{2}......(2)$
We know that value of $\sin x$lies between 1 and -1
So, $\sin x = 2$is not possible
The value of $\sin x = - \dfrac{1}{2}$ is the solution of the equation.
And $\sin \dfrac{\pi }{6} = \sin 30^\circ = \dfrac{1}{2}$
We know that we need a negative angle of sin, which can be in third and fourth quadrants.
So, we add $\pi $ to$\dfrac{\pi }{6}$and subtract$\dfrac{\pi }{6}$ from $2\pi $,as we want sin in third and fourth quadrants.
We can write $\sin \left( {\dfrac{\pi }{6} + \pi } \right) = \sin \dfrac{{7\pi }}{6} = - \dfrac{1}{2}......(3)$
And $\sin \left( {2\pi - \dfrac{\pi }{6}} \right) = \sin \dfrac{{11\pi }}{6} = - \dfrac{1}{2}......(4)$
Now, we equate equation 2 with equation 3 and 4
$\sin x = \sin \dfrac{{7\pi }}{6}$ And$\sin x = \sin \dfrac{{11\pi }}{6}$
$\sin $will cancel out from both sides in both above equations.

Hence, we got value of $x = \dfrac{{7\pi }}{6}$and $x = \dfrac{{11\pi }}{6}$

Note:
We should keep in mind that in the trigonometry equation firstly we should apply a suitable identity to simplify the equation as per our requirement, and after simplification we automatically get the way to solve the simplified equation easily.