Question

How do you solve $\cos (2x) = \dfrac{1}{2}$ , and find all exact general solutions $?$

Answer 1

We will equate both the equation using the value of $\cos$ function and, then we will try to find the solution for $x$ . On doing some calculation we get the required answer.

Formula Used:
Cosine is a periodic function.
So, a cosine function can have multiple solutions.
Since the periodic function of Cosine is $\cos (\theta + 2n\pi ) = \cos (\theta )$ , general solutions for Cosine function is as following:
$\theta + 2n\pi$ , for all values of $\theta$ and $n \in N$ , where $N$ is a natural number.
We can clearly say that $2\pi$ is the smallest possible positive real number for all possible values of $\theta + 2n\pi$ .
So, $\cos (\theta + 2n\pi ) = \cos (\theta )$ has an equal period of $2\pi$ .
We also know that if the value of $n$ is a positive or negative whole number then $\cos (2n\pi \pm \theta ) = \cos (\theta )$ .

Complete step by step answer:
We have to solve for $x$ in the above given equation of $\cos (2x) = \dfrac{1}{2}$ .
We know that the value of $\cos \left( {\dfrac{\pi }{3}} \right)$ is equal to $\dfrac{1}{2}$ .
So, we can re-write the above equation in following way:
$\Rightarrow \cos (2x) = \cos (2n\pi + \dfrac{\pi }{3})$ .
So, if we cancel the cosine function from both of the sides, we get:
$\Rightarrow 2x = 2n\pi + \dfrac{\pi }{3}$ .
Now, if we divide both the sides by $2$ , we get:
$\Rightarrow x = n\pi + \dfrac{\pi }{6}$ , for all the values of $n \in N$ .
Now, the value of cosine is always positive in the first quadrant and in the fourth quadrant.
So, to find the value of $\theta$ , we need to subtract the reference angle from $2\pi$ , so that we could have the value of $\theta$ in the fourth quadrant.
So, the value of $\theta$ will become $\left( {2\pi - \dfrac{\pi }{3}} \right) = \left( {\dfrac{{6\pi - \pi }}{3}} \right) = \dfrac{{5\pi }}{3}.$
So, we can re-write the general equation of cosine as following:
$\Rightarrow \cos (2x) = \cos \left( {2n\pi + \dfrac{{5\pi }}{3}} \right)$ .
So, we can re-write it as following:
$\Rightarrow 2x = 2n\pi + \dfrac{{5\pi }}{3}$ .
Now, divide both the sides by $2$ , we get:
$\Rightarrow x = n\pi + \dfrac{{5\pi }}{6}$ , for all the values of $n \in N$ .

$\therefore x = \left( {n\pi + \dfrac{\pi }{6}} \right),\left( {n\pi + \dfrac{{5\pi }}{6}} \right)$ , for all integer value of $n$ .

Note: Points to remember:
For all the integer value of $n$ , the value of $\cos \left( {n{{90}^ \circ } \pm \theta } \right)$ is always equals to $\cos (\theta )$
Both the Sine and Cosine curve have the same period of $2\pi$ .