How do you find the general solution for ${{\sin }^{2}}x=3{{\cos }^{2}}x$ ?

Hint: In this question, we have to find the value of x of an equation. The equation given to us consists of trigonometric functions. Therefore, we will apply the trigonometric formulas to get the solution for the answer. We begin this problem by dividing both sides of the equation by ${{\cos }^{2}}x$ and then apply the trigonometric formula $\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}={{\tan }^{2}}x$ on the left-hand side of the equation and make further simplification. Then, we will take the square root of both sides of the equation, and therefore, we get 2 equations. So, we will solve them separately to get the solution to the problem.

Complete step by step answer:
According to the problem, we have to find the value of x.
The equation given to us is ${{\sin }^{2}}x=3{{\cos }^{2}}x$ ---------- (1)
Thus, we will apply the trigonometric formulas to get the solution.
We first divide ${{\cos }^{2}}x$on both sides in the equation (1), we get
$\Rightarrow \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}=\dfrac{3{{\cos }^{2}}x}{{{\cos }^{2}}x}$
Thus, we will now apply the trigonometric formula $\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}={{\tan }^{2}}x$ on the left-hand side of the above equation, we get
$\Rightarrow {{\tan }^{2}}x=\dfrac{3{{\cos }^{2}}x}{{{\cos }^{2}}x}$
On further solving, we get
$\Rightarrow {{\tan }^{2}}x=3$
Now, we will take the square root on both sides of the equation, we get
$\Rightarrow \sqrt{{{\tan }^{2}}x}=\sqrt{3}$
Thus, we get
$\Rightarrow \tan x=\pm \sqrt{3}$
Therefore, from the above equation, we get two separate equations, that is
$\tan x=\sqrt{3}$ ------ (2) and
$\tan x=-\sqrt{3}$ ----------- (3)
Now, we will solve equation (2), which is
$\tan x=\sqrt{3}$
Thus, we take ${{\tan }^{-1}}$ on both sides of the above equation, we get
${{\tan }^{-1}}\left( \tan x \right)={{\tan }^{-1}}\sqrt{\left( 3 \right)}$
As we know, the trigonometric formula ${{\tan }^{-1}}\left( \tan a \right)=a$ , therefore the above equation will become
$x={{\tan }^{-1}}\sqrt{\left( 3 \right)}$
$x=\dfrac{\pi }{3}+n\pi $ where n is some integer
Now, we will solve equation (3), which is
$\tan x=-\sqrt{3}$
Thus, we take ${{\tan }^{-1}}$ on both sides of the above equation, we get
${{\tan }^{-1}}\left( \tan x \right)={{\tan }^{-1}}\left( -\sqrt{\left( 3 \right)} \right)$
As we know, the trigonometric formula ${{\tan }^{-1}}\left( \tan a \right)=a$ , therefore the above equation will become
$x={{\tan }^{-1}}\left( -\sqrt{\left( 3 \right)} \right)$
$x=\dfrac{2\pi }{3}+n\pi $ where n is some integer
Therefore, for the trigonometric equation ${{\sin }^{2}}x=3{{\cos }^{2}}x$ , its general solution is the value of x, that is either $x=\dfrac{\pi }{3}+n\pi $ or $x=\dfrac{2\pi }{3}+n\pi $ , where n is some integer.

Note: While solving this problem, do mention all the trigonometric formulas in the steps wherever they are applicable. Keep in mind that the tan function is a periodic function that implies its value repeats after some interval, that is why we add $n\pi $ to the values of x. It means that $n\pi $ is added after every interval.