How do you solve ${\cos ^2}x - \cos x = 0$ ?

Hint: In the question, a polynomial equation of one variable is given, it is a function of $\cos x$ . To make it easier to solve the given polynomial equation, we replace $\cos x$ in the given equation with any other variable. We know that the degree of a polynomial equation is the highest power of the variable used in the polynomial and also the number of roots of a polynomial equation is equal to its degree, so several methods like factorizing the equation or by a special formula called quadratic formula or by completing the square method or quadratic formula can be used to solve the given equation; we use other methods if we are not able to factorize the equation.

Complete step by step answer:
We have to solve the equation ${\cos ^2}x - \cos x = 0$
Taking $\cos x = t$, we get –
$ \Rightarrow {t^2} - t = 0$
Now we see that this equation can be solved by simply taking t as common –
$
 \Rightarrow t(t - 1) = 0 \\
   \Rightarrow t = 0,\,t - 1 = 0 \\
   \Rightarrow t = 0,\,t = 1 \\
 $
Putting the original value of t as $\cos x$ , we get –
$ \Rightarrow \cos x = 0,\,\cos x = 1$
We know that $\cos 0 = 0,\,\cos 90^\circ = 1$, so we get –
$
 \Rightarrow \cos x = \cos 0,\,\cos x = \cos 90^\circ \\
   \Rightarrow x = 0^\circ ,\,x = 90^\circ \\
 $
Hence, the solution of the equation ${\cos ^2}x - \cos x = 0$ is $x = 0^\circ $ or $x = 90^\circ $.

Note: In this question, we have just written the general solution, there can be infinitely many solutions to the questions like this. The general solution lies in the interval $[0,2\pi )$ that is the value of x can be greater than or equal to zero but smaller than $2\pi $ so we take only these two values as the answer.