How do you solve the following equation $6{\sin ^2}x - \sin x - 2 = 0$ in the interval $\left[ {0,2\pi} \right]$?

Hint: In this problem we have given some trigonometric equation and which lies in some interval. And we are asked to solve the given trigonometric equation in the given interval. For solving this equation first we are going to change the given equation as a quadratic equation and then we proceed.

Complete step-by-step solution:
Given equation is $6{\sin ^2}x - \sin x - 2 = 0$ in the interval $\left[ {0,2\pi} \right]$.
First we are going to replace the trigonometric terms from the given equation by substituting some value, we get
Let’s use the substitution $\sin x = u$, then our equation becomes
$6{u^2} - u - 2 = 0$
Now we have a quadratic equation and also we know how to solve it.
Let’s factor the quadratic equation.
Now the middle term of a quadratic equation $ - u$ can be written as $ - 4u + 3u$.
$ \Rightarrow 6{u^2} - 4u + 3u - 2 = 0$
We can factor out a $3u$ from the first pair and a $ - 2$ from the second pair, we get
$3u\left( {2u + 1} \right) - 2\left( {2u + 1} \right) = 0$, here the common factor is $\left( {2u + 1} \right)$ and let’s take it out, then
$ \Rightarrow \left( {2u + 1} \right)\left( {3u - 2} \right) = 0$
Using the zero product property, we can set two factors equal to $0$ and solve
$2u + 1 = 0$ and $3u - 2 = 0$
$ \Rightarrow u = - \dfrac{1}{2}$ and $u = \dfrac{2}{3}$
Now remember that $\sin x = u$, so
$\sin x = - \dfrac{1}{2}$ and $\sin x = \dfrac{2}{3}$
Now we need to solve this on $\left[ {0,2\pi} \right]$.
When $\sin x = - \dfrac{1}{2}$, the unit circle tells us that $\sin x = - \dfrac{1}{2}$ when $x = \dfrac{{7\pi }}{6}$ and $x = \dfrac{{11\pi }}{6}$, so these are two solutions.
When $\sin x = \dfrac{2}{3}$, taking inverse sine on both sides, we get
$ \Rightarrow {\sin ^{ - 1}}(\sin x) = {\sin ^{ - 1}}\left( {\dfrac{2}{3}} \right)$
$ \Rightarrow x = {\sin ^{ - 1}}\left( {\dfrac{2}{3}} \right)$
Using the calculator, we obtain the principal solution $x \approx 0.73$ and also there is another solution which is obtained by subtracting $0.73$ from $\pi $ that is $x = \pi - 0.73 \approx 2.41$.

Therefore the solutions are $x = \dfrac{{7\pi }}{6},x = \dfrac{{11\pi }}{6},x = 0.73$ and $x = 2.41$

Note: When $\sin x = - \dfrac{1}{2}$ we got the values for $x$ as $x = \dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}$. For, we know that $\sin x$ is negative in the third and fourth quadrants.
Since $\dfrac{{\sin \pi }}{6} = \dfrac{1}{2}$, $\sin x$ would be $ - \dfrac{1}{2}$ for $x = \pi + \dfrac{\pi }{6}$ for $x = 2\pi - \dfrac{\pi }{6}$. When $\sin x = \dfrac{2}{3}$ using the calculator we got $x \approx 0.73$.
However, there is another solution that the calculator won’t give us. To find the second answer we subtracted $0.73$ from $\pi $ and we get $x = 2.41$.