Question

How do you solve $3{{\tan }^{2}}\theta =1$?

Answer 1

In order to solve any trigonometric equation, we have to follow three steps. First step is to find all the values of the trigonometric ratios in the equation needed to solve the equation. Second step is to find all angles that give us these values from the first step. And the last step is to get a common relation or the general solution of the equation.

Complete step by step answer:
As per the given question, we are provided with a trigonometric equation. That trigonometric equation is $3{{\tan }^{2}}\theta =1$. We can rewrite the given trigonometric equation as
$\Rightarrow {{\tan }^{2}}\theta =\dfrac{1}{3}$
Here, we have a square on $\tan \theta$. So, we need to take square root on both sides of the equation. Then on doing so, we get
$\Rightarrow \sqrt{{{\tan }^{2}}\theta }=\sqrt{\dfrac{1}{3}}$
We know that the square root of a square term gives both positive and negative terms. So, we get
$\Rightarrow \pm \tan \theta =\dfrac{1}{\sqrt{3}}$
We can rewrite the equation as
$\Rightarrow \tan \theta =\pm \dfrac{1}{\sqrt{3}}$
Hence, we can say that $\tan \theta$ has two values $\dfrac{1}{\sqrt{3}}$ and $-\dfrac{1}{\sqrt{3}}$.
Here, we have two cases. In case (1), $\tan \theta$ has the value $\dfrac{1}{\sqrt{3}}$. In case (2), $\tan \theta$ has the value $-\dfrac{1}{\sqrt{3}}$.
Let us consider case (1) where $\tan \theta =\dfrac{1}{\sqrt{3}}$. In the interval $\left[ 0,2\pi \right]$, $\tan \theta$ is positive in the first and third quadrant. In the first quadrant, $\tan \theta =\dfrac{1}{\sqrt{3}}$ for $\theta =\dfrac{\pi }{6}$. In the third quadrant, $\tan \theta =\dfrac{1}{\sqrt{3}}$ for $\theta =\dfrac{7\pi }{6}$. We can rewrite $\theta =\dfrac{7\pi }{6}$ as $\theta =\pi +\dfrac{\pi }{6}$. Here, we
can generalize the solution as $\tan \theta =\dfrac{1}{\sqrt{3}}$ for $\theta =\dfrac{\pi }{6}+k\pi$, where $k\in \mathbb{Z}$.

Now, let us consider case (2) where $\tan \theta =-\dfrac{1}{\sqrt{3}}$. In the interval $\left[ 0,2\pi \right]$, $\tan \theta$ is negative in the second and fourth quadrant. In the second quadrant, $\tan \theta =-\dfrac{1}{\sqrt{3}}$
for $\theta =\dfrac{5\pi }{6}$. In the fourth quadrant, $\tan \theta =-\dfrac{1}{\sqrt{3}}$ for $\theta =\dfrac{11\pi }{6}$. Here, we can generalize the solution as $\tan \theta =-\dfrac{1}{\sqrt{3}}$ for $\theta =-\dfrac{\pi }{6}+k\pi$, where $k\in \mathbb{Z}$.

$\therefore \theta =\pm \dfrac{\pi }{6}+k\pi$ are the two solutions of the equation $3{{\tan }^{2}}\theta =1$.

Note: While solving the trigonometric equations, we have to be very careful about the angles satisfying the required conditions. We need to remember that the general solution must give all the solutions. We need to convert the angles properly.