Question

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

Answer 1

Hint: 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 \[2\pi \] .