How do you find all the solutions in the interval $[0,2\pi )$ without using a calculator of $3\sin x = 2{\cos ^2}x$ ?

Hint: Firstly we convert the given equation into a quadratic equation which will be in the form $a{y^2} + by + c = 0$ where $y$ will be $\sin x$ . Then we will get the value of $\sin x$ from the quadratic equation and then we will find the value of $x$ where $x$ lies in the interval $[0,2\pi )$ .

Formulas Used:
$\Rightarrow$${\cos ^2}x + {\sin ^2}x = 1$
$\Rightarrow$$\sin \theta = \sin z$
$ \Rightarrow \theta = 2n\pi + z$
where $n = 0, \pm 1, \pm 2,.....$

Complete step-by-step answer:
$\Rightarrow$$3\sin x = 2{\cos ^2}x$
As we know that ;
$\Rightarrow$${\cos ^2}x + {\sin ^2}x = 1 \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x$
Applying this in the equation we will get;
$\Rightarrow$$3\sin x = 2(1 - {\sin ^2}x)$
$ \Rightarrow 3\sin x = 2 - 2{\sin ^2}x$
Adding both sides $2{\sin ^2}x$ and subtracting $2$ from both side we will get;
$ \Rightarrow 2{\sin ^2}x + 3\sin x - 2 = 0$
$3\sin x$ can be written as $(4 - 1)\sin x$.
$ \Rightarrow 2{\sin ^2}x + (4 - 1)\sin x - 2 = 0$
$ \Rightarrow 2{\sin ^2}x + 4\sin x - \sin x - 2 = 0$
Taking $2\sin x$ common from the first two terms and $ - 1$ from the next two terms we will get;
$ \Rightarrow 2\sin x(\sin x + 2) - (\sin x + 2) = 0$
$ \Rightarrow (2\sin x - 1)(\sin x + 2) = 0$
We will obtain two cases from here;
Case $1$ :
$\Rightarrow$$2\sin x - 1 = 0$
$ \Rightarrow 2\sin x = 1$
$ \Rightarrow \sin x = \dfrac{1}{2}$
Case $2$ :
$\Rightarrow$$\sin x + 2 = 0$
$ \Rightarrow \sin x = - 2$
In case $2$ we can see that $\sin x = - 2$ . But we know that the minimum value of $\sin x$ is $ - 1$ .
So $\sin x = - 2$ is impossible.
From case $1$ ;
When $x$ lies in the interval $\left[ {0,\dfrac{\pi }{2}} \right]$ ;
$ \Rightarrow \sin x = \dfrac{1}{2}$
$ \Rightarrow \sin x = \sin \dfrac{\pi }{6}$
Now using the formula;
$\sin \theta = \sin z$
$ \Rightarrow \theta = 2n\pi + z$
where $n = 0, \pm 1, \pm 2,.......$
$ \Rightarrow x = 2n\pi + \dfrac{\pi }{6}$
Now $x$ lies in the interval $[0,2\pi )$ so the only possibility is $n = 0$ ;
$\therefore x = \dfrac{\pi }{6}$ .
When $x$ lies in the interval $\left[ {\dfrac{\pi }{2},\pi } \right]$ ;
$ \Rightarrow \sin x = \dfrac{1}{2}$
$ \Rightarrow \sin x = \sin \dfrac{{5\pi }}{6}$
Now using the formula;
$\Rightarrow$$\sin \theta = \sin z$
$ \Rightarrow \theta = 2n\pi + z$
Where $n = 0, \pm 1, \pm 2,.......$
$ \Rightarrow x = 2n\pi + \dfrac{{5\pi }}{6}$
Now $x$ lies in the interval $[0,2\pi )$ so the only possibility is $n = 0$ ;
$\therefore x = \dfrac{{5\pi }}{6}$

So the values of $x$ are $\dfrac{\pi }{6}$ and $\dfrac{{5\pi }}{6}$.

Alternative method:
We can also deduce the equation in the form of $\cos x$ by putting $\sin x = \sqrt {1 - {{\cos }^2}x} $.

Note:
When the interval changes the trigonometric functions like $\sin $ , $\cos $ , $\tan $ etc. changes the values of $x$ . $\sin $ function gives positive values of $x$ in the intervals $\left[ {0,\dfrac{\pi }{2}} \right]$ and $\left[ {\dfrac{\pi }{2},\pi } \right]$ which are known as the first quadrant and second quadrant respectively. $\sin $ gives negative values of $x$ in the intervals $\left[ {\pi ,\dfrac{{3\pi }}{2}} \right]$ and $\left[ {\dfrac{{3\pi }}{2},2\pi } \right]$ which are known as third and fourth quadrant respectively.