How do you solve ${{\cos }^{3}}\left( x \right)=\cos \left( x \right)$ on the interval of $[0,2\pi )?$

Hint: Here a trigonometric equation is given with its interval which we have to simplify it.
Here, We are simply substituting the other variable $'u'$ at the place of $\cos x$
We did it to make the equation more simple.
After simplifying it we will get the value for the variable $'u'$ which is nothing but the value of $\cos x$ put that all, values in $\cos x$ and we will get the final answer in terms of angles.

Complete step by step solution:
Given that, there is an equation,
${{\cos }^{3}}\left( x \right)=\cos \left( x \right)...(i)$
Whose interval is $[0,2\pi )$ and we have to solve it. Here, we have to use a substitution method that is putting a variable at the place of $\cos \left( x \right)$ and then we have to solve it in the way of a normal regular polynomial.
${{\cos }^{3}}\left( x \right)=\cos \left( x \right)$
${{\left( \cos x \right)}^{3}}=\cos \left( x \right)$
Putting $\cos x=u$ in above equation, we get
${{u}^{3}}=u$
Transpose $u$ at the left side to form an equation.
${{u}^{3}}-u=0$
$u\left( {{u}^{2}}-1 \right)=0$
$u\left( u-1 \right)\left( u+1 \right)=0$
Separate the above terms, such that,
$u=0$
$u-1=0$, $u+1=0$
$u=1$, $u=-1$
The values of $u=0,1,-1$
$\cos x=0,1,-1$
Now, put this value in $\cos x$ i.e. in cosine equation i.e. put $\cos x=0$
$x={{\cos }^{-1}}\left( 0 \right)$
$x=\dfrac{\pi }{2},\dfrac{3\pi }{2}$
Now, put $\cos x=1$
$x={{\cos }^{-1}}\left( 1 \right)$
$x=0$
Now, put $\cos x=1$
$x={{\cos }^{-1}}\left( 1 \right)$
$x=0$
Now, put $\cos x=-1$
$x={{\cos }^{-1}}\left( -1 \right)$
$x=\pi $

Therefore, the final answer for the given trigonometric equation is $x=0,\pi ,\dfrac{\pi }{2},\dfrac{3\pi }{2}$.

Note: In this question, the trigonometric equation is given which we have to solve.
Here we put $\cos x=u$ just for simplifying the equation. Otherwise using trigonometric functions the equation may become more complicated.
The final answer we got is nothing but the trigonometric cosine values in degree.
i.e. ${{\cos }^{-1}}\left( 0 \right)=\dfrac{\pi }{2},\dfrac{3\pi }{2}$ i.e. $90{}^\circ ,270{}^\circ $
${{\cos }^{-1}}\left( 1 \right)=0$ i.e. $0{}^\circ $
${{\cos }^{-1}}\left( -1 \right)\pi $ i.e. $180{}^\circ $