Question

How do you solve ${\cos ^2}x = \,\dfrac{3}{4}\,$?

Answer 1

Hint : In this question first, we have the square root on both sides and then we have to find the angle whose cosine is equal to the square root of the value given in the right hand side. You can think that there is only one value but there can be multiple values of angle.

Complete step-by-step answer :
In this question, it is given that ${\cos ^2}x = \,\dfrac{3}{4}\,$
Now taking the root on both sides, we get
$\Rightarrow \,\cos x = \pm \dfrac{{\sqrt 3 }}{2}$
Now, taking the inverse on both sides
$\Rightarrow x = co{s^{ - 1}} \pm \dfrac{{\sqrt 3 }}{2}$
Now, there are two cases
Case I:
$\Rightarrow {\cos ^{ - 1}}\dfrac{{\sqrt 3 }}{2} =$ An angle whose cosine is equal to $\dfrac{{\sqrt 3 }}{2}$.
Hence, $x\, = \,2n\pi \, \pm \dfrac{\pi }{6}$ where $n \in Z\,,\,Z$ represents the set of all integers.
Case II:
$\Rightarrow {\cos ^{ - 1}} - \dfrac{{\sqrt 3 }}{2} =$An angle whose cosine is equal to $- \dfrac{{\sqrt 3 }}{2}$.
Hence, $x\, = \,2n\pi \, \pm \,\dfrac{{5\pi }}{6}$ where $n \in Z\,,\,Z$ represents the set of all integers.
This means that after every one rotation we get the same value of $\cos x$.
Therefore, the solution to expression ${\cos ^2}x = \,\dfrac{3}{4}\,$ is $\,2n\pi \, \pm \dfrac{\pi }{6}\,and\,\,2n\pi \, \pm \,\dfrac{{5\pi }}{6}$ where $n \in Z\,,\,Z$ represents the set of all integers.

Note : There are multiple ways to do this question. We can also convert the ${\cos ^2}x$ into ${\sec ^2}x$ by using formulas and then ${\sec ^2}x$ into ${\tan ^2}x$ to solve this problem. We can also do this question by converting ${\cos ^2}x$ in terms of ${\sin ^2}x$ and then we can solve the equation. It is a question of trigonometric equations. Always be careful while doing the questions of trigonometric equations because there can be multiple solutions of a question. Sometimes we can also do the questions using graphs.