How do you solve $2{\cos ^2}x = 1$?

Hint: Cosine is a trigonometric function. It is the ratio of base and hypotenuse of the right-angled triangle. All trigonometric functions are periodic, that is, they give the same output if their period is added or subtracted in their input. It can be easily converted into a sine function for ease of calculation, and this property of the cosine function is very useful for us in this question.

Complete step by step solution:
According to the question we have to find the value of $x$ from the equation $2{\cos ^2}x = 1$
So,
$ \Rightarrow {\cos ^2}x = \dfrac{1}{2}$
$ \Rightarrow \cos x = \pm \sqrt {\dfrac{1}{2}} $
We know the value of $\cos \theta $ for different values of $\theta $ ($\theta $ is input for the function)
So we know the value of $\cos x$ when we will get the required value
Hence, we get our $x$ equal to $\dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4}$ and $\dfrac{{7\pi }}{4}$
We can also solve it by changing cosine function to sine function,
$ \Rightarrow 2(1 - {\sin ^2}x) = 1$ (Convert cos to sin)
$ \Rightarrow 2 - 2{\sin ^2}x = 1$
$ \Rightarrow 2{\sin ^2}x = 1$
$ \Rightarrow {\sin ^2}x = \dfrac{1}{2}$
$ \Rightarrow \sin x = \pm \sqrt {\dfrac{1}{2}} $
We know the value of $\sin \theta $ for different values of $\theta $ ($\theta $ is input for the function)
So we know that the value of $\sin x$ when we will get the required value
Hence, we get our $x$ equal to $\dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4}$ and $\dfrac{{7\pi }}{4}$
Hence, these values of x are our answer.

Note: We can clearly observe how easily we can change the cosine function into sine function for our use in the question. These properties are very helpful in many cases and they must be remembered. Taking both positive and negative signs while square rooting the number is also to be remembered. If we miss it, we will get only two solutions instead of four solutions.