Question

How do you solve $\tan x = - \sqrt 3$ ?

Answer 1

Hint : This is a question of trigonometric identities. There can be multiple values of x in this question. We just need to find in which quadrant these values lie and accordingly we have to give the values of x. We should also keep in mind that $\tan x$ is a periodic function with a period of $\pi$. So, accordingly we will find the values of x.

Complete step-by-step answer :
In the given question, we have
$\Rightarrow \tan x = - \sqrt 3$
We know that the value of $\tan x$ is negative in two quadrants, that is the second quadrant and fourth quadrant.
We have a general formula for $\tan x = \tan\theta$ , then $x = n\pi + \theta$
Therefore, the value of x in second quadrant is $x = \dfrac{{2\pi }}{3}$ and the value of x in fourth quadrant is $x = \dfrac{{5\pi }}{6}$ for $\tan x$ to be equals to $- \sqrt 3$ .
The period of the $\tan x$ function is $\pi$ so values will repeat every $\pi$radian in both directions.
Therefore, $x = \dfrac{{2\pi }}{3} + n\pi$ ,$\dfrac{{5\pi }}{6} + n\pi$ , for any integer n.
So, the correct answer is “, $x = \dfrac{{2\pi }}{3} + n\pi$ ,$\dfrac{{5\pi }}{6} + n\pi$ ”.

Note : Angles are measured by rotating clockwise from the positive x-axis. Generally, whenever a value of a trigonometric function is given, there are multiple angles involved with it. As $\tan x$, $\sin x$ and $\cos x$ are periodic functions they repeat their values after a regular interval of time. Like $\sin x$ and $\cos x$ are periodic functions with a period of $2\pi$ . This means that their values repeat after a regular interval of $2\pi$. Also, $\tan x$ is a periodic function with a period $\pi$ . It means that its value repeats after a regular interval of $\pi$.