Question

How do you find all the solutions to the equation \[\cos \left( 2x \right)=\dfrac{1}{2}\]?

Answer 1

Hint: We can solve this question by using trigonometric identities. As cos is a periodic function we can expect many solutions. First we will write the RHS as cos angle and then we solve the problem by using the solutions we have for that equation. By solving those solutions we have we will get the solution.

Complete step-by-step solution:
The General solution for \[\cos x=\cos y\] is \[x=2n\pi \pm y\]
We will use this solution to solve the problem.
Given
\[\cos \left( 2x \right)=\dfrac{1}{2}\]
We have to write the RHS as a cos function. We know that \[\dfrac{1}{2}\] as an angle of \[{{60}^{\circ }}\]. So we can write \[\dfrac{1}{2}\] as \[\dfrac{\pi }{3}\].
We can write \[\dfrac{1}{2}\] as \[\cos \left( \dfrac{\pi }{3} \right)\].
So we have to substitute \[\cos \left( \dfrac{\pi }{3} \right)\] in the equation. We will get
\[\Rightarrow \cos \left( 2x \right)=\cos \left( \dfrac{\pi }{3} \right)\]
As said above the general solution for \[\cos x=\cos y\] is \[x=2n\pi \pm y\].
Using this we can write the solution for our equation.
We will get
\[2x=2n\pi \pm \dfrac{\pi }{3}\]
To get the value of x we have to divide it by 2.
We will get
\[\Rightarrow \dfrac{2x}{2}=\dfrac{2n\pi \pm \dfrac{\pi }{3}}{2}\]
By simplifying it we will get
\[\Rightarrow x=\dfrac{\pi }{6}\pm n\pi \]
So the solution for the given equation are
\[x=\dfrac{\pi }{6}\pm n\pi \]
As cos is a periodic equation we may have many solutions. The solution can also be like
 \[2x=2n\pi \pm \dfrac{\pi }{3}\].
Like this we can write a few more solutions.
The solutions are \[x=\dfrac{\pi }{6}\pm n\pi \] and \[2x=2n\pi \pm \dfrac{\pi }{3}\].

Note: We should be aware of the general solutions of the function. Otherwise we cannot solve the question. And also we should be careful while creating periodic functions solutions because there is a chance we forgot to write period correctly.