Question

How do you find the solution to $3{\tan ^2}\theta = 1$ if $0 \leqslant \theta < {360^ \circ }$?

Answer 1

First, divide both sides of the equation by $3$ and take square root on both sides of the equation, Then, find the values of $\theta$ satisfying $\tan \theta = \dfrac{1}{{\sqrt 3 }}$ using trigonometric properties.
Next, find the values of $\theta$ satisfying $\tan \theta = - \dfrac{1}{{\sqrt 3 }}$ using trigonometric properties. Next, find all values of $\theta$ in the interval $0 \leqslant \theta < {360^ \circ }$. Then, we will get all solutions of the given equation in the given interval.

Formula used:
$\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}$
$\tan \left( {\pi - x} \right) = - \tan x$
$\tan \left( {2\pi - x} \right) = - \tan x$

Complete step by step answer:
Given equation: $3{\tan ^2}\theta = 1$
We have to find all possible values of $\theta$ satisfying given equation in the interval $0 \leqslant \theta < {360^ \circ }$.
Divide both sides of the equation by $3$, we get
${\tan ^2}\theta = \dfrac{1}{3}$
Take square root on both sides of the equation, we get
$\tan \theta = \pm \dfrac{1}{{\sqrt 3 }}$
First, we will find the values of $\theta$ satisfying $\tan \theta = \dfrac{1}{{\sqrt 3 }}$.
So, take the inverse tan of both sides of the equation to extract $\theta$ from inside the tan.
$\theta = \arctan \left( {\dfrac{1}{{\sqrt 3 }}} \right)$
Since, the exact value of $\arctan \left( {\dfrac{1}{{\sqrt 3 }}} \right) = \dfrac{\pi }{6}$.
$\Rightarrow \theta = \dfrac{\pi }{6}$
Since, the tan function is positive in the first and third quadrants.
So, to find the second solution, add the reference angle from $\pi$ to find the solution in the fourth quadrant.
$\theta = \pi + \dfrac{\pi }{6}$
$\Rightarrow \theta = \dfrac{{7\pi }}{6}$
Since, the period of the $\tan \theta$ function is $\pi$ so values will repeat every $\pi$ radians in both directions.
$\theta = \dfrac{\pi }{6} + n\pi ,\dfrac{{7\pi }}{6} + n\pi$, for any integer $n$.
Now, we will find the values of $\theta$ satisfying $\tan \theta = - \dfrac{1}{{\sqrt 3 }}$…(i)
So, using the property $\tan \left( {\pi - x} \right) = - \tan x$ and $\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}$ in equation (i).
$\Rightarrow \tan \theta = - \tan \dfrac{\pi }{6}$
$\Rightarrow \tan \theta = \tan \left( {\pi - \dfrac{\pi }{6}} \right)$
$\Rightarrow \theta = \dfrac{{5\pi }}{6}$
Now, using the property $\tan \left( {2\pi - x} \right) = - \tan x$ and $\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}$ in equation (i).
$\Rightarrow \tan \theta = - \tan \dfrac{\pi }{6}$
$\Rightarrow \tan \theta = \tan \left( {2\pi - \dfrac{\pi }{6}} \right)$
$\Rightarrow \theta = \dfrac{{11\pi }}{6}$
Since, the period of the $\tan \theta$ function is $\pi$ so values will repeat every $\pi$ radians in both directions.
$\theta = \dfrac{{5\pi }}{6} + n\pi ,\dfrac{{11\pi }}{6} + n\pi$, for any integer $n$.
Now, find all values of $\theta$ in the interval $0 \leqslant \theta < {360^ \circ }$.
Since, it is given that $\theta \in \left[ {0,{{360}^ \circ }} \right)$, hence put $n = 0$ in the general solution.
So, putting $n = 0$ in $\theta = \dfrac{\pi }{6} + n\pi ,\dfrac{{7\pi }}{6} + n\pi$, we get
$\theta = \dfrac{\pi }{6},\dfrac{{7\pi }}{6}$
Now, putting $n = 0$ in $\theta = \dfrac{{5\pi }}{6} + n\pi ,\dfrac{{11\pi }}{6} + n\pi$,we get
$\theta = \dfrac{{5\pi }}{6},\dfrac{{11\pi }}{6}$
Thus, $\theta = \dfrac{\pi }{6},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}$ or $\theta = {30^ \circ },{150^ \circ },{210^ \circ },{330^ \circ }$.

Hence, $\theta = \dfrac{\pi }{6},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}$ or $\theta = {30^ \circ },{150^ \circ },{210^ \circ },{330^ \circ }$ are solutions of the given equation in the interval $0 \leqslant \theta < {360^ \circ }$.

Note: In above question, we can find the solutions of given equation by plotting the equation, $3{\tan ^2}\theta = 1$ on graph paper and determine all solutions which lie in the interval, $0 \leqslant \theta < {360^ \circ }$.

From the graph paper, we can see that there are four values of $\theta$ in the interval $0 \leqslant \theta < {360^ \circ }$.
So, these will be the solutions of the given equation in the given interval.
Final solution: Hence, $\theta = \dfrac{\pi }{6},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}$ or $\theta = {30^ \circ },{150^ \circ },{210^ \circ },{330^ \circ }$ are solutions of the given equation in the interval $0 \leqslant \theta < {360^ \circ }$.