Question

How do you solve ${{\tan }^{2}}\left( 3x \right)=3$ and find all exact general solutions?

Answer 1

Hint: We solve the root value for ratio tan. We explain the function $arc\tan \left( x \right)$. We express the inverse function of tan in the form of $arc\tan \left( x \right)={{\tan }^{-1}}x$. We draw the graph of $arc\tan \left( x \right)$ and the line $x=\pm \sqrt{3}$ to find the intersection point as the solution.

Complete step by step answer:
The given trigonometric equation is ${{\tan }^{2}}\left( 3x \right)=3$. We take the root square on both sides and get $\tan \left( 3x \right)=\pm \sqrt{3}$.
The solution of x is the inverse function of trigonometric ratio tan.
The arcus function represents the angle which on ratio tan gives the value.
So, $arc\tan \left( x \right)={{\tan }^{-1}}x$. If $arc\tan \left( x \right)=\alpha $ then we can say $\tan \alpha =x$.
Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of $2\pi $.
The general solution for that value where $\tan \alpha =x$ will be $n\pi +\alpha ,n\in \mathbb{Z}$.
We first use the principal value. For ratio tan we have $-\dfrac{\pi }{2}\le arc\tan \left( x \right)\le \dfrac{\pi }{2}$.
Now we take the function as $y=\tan \left( 3x \right)=\pm \sqrt{3}$. The graph of the function $y=\tan \left( 3x \right)$ is

Let the angle be $\theta $ for which $arc\tan \left( x \right)={{\tan }^{-1}}x=\theta $. This gives $\tan \left( 3x \right)=\pm \sqrt{3}$.
We know that \[\tan \left( 3x \right)=\pm \sqrt{3}=\tan \left( \pm \dfrac{\pi }{3} \right)\] which gives $\theta =\pm \dfrac{\pi }{3}$. For this we take the line of $y=\pm \sqrt{3}$ and see the intersection of the line with the graph $arc\tan \left( x \right)$.

The general solution of the function $arc\tan \left( x \right)$ is $n\pi +\alpha ,n\in \mathbb{Z}$
The general solution of the function $\tan \left( 3x \right)=\pm \sqrt{3}$ is $3x=n\pi \pm \dfrac{\pi }{3},n\in \mathbb{Z}$. The simplified solution for ${{\tan }^{2}}\left( 3x \right)=3$ is \[x=\left( 3n\pm 1 \right)\dfrac{\pi }{9},n\in \mathbb{Z}\].

Note: If we are finding an $arc\tan \left( x \right)$ of a positive value, the answer is between $0\le arc\tan \left( x \right)\le \dfrac{\pi }{2}$. If we are finding the $arc\tan \left( x \right)$ of a negative value, the answer is between $-\dfrac{\pi }{2}\le arc\tan \left( x \right)\le 0$.