Question

How do you solve $3{{\tan }^{2}}x-1=0$ for ${{0}^{\circ }}\le x\le {{360}^{\circ }}$ ? $$$$

Answer 1

We solve for $\tan x$ from the given equation by dividing both sides by 3 and then factoring with identity${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. We find the principal solution $x=\alpha$ from trigonometric table and then find general solution of $\tan \theta =\tan \alpha$ as $\theta =n\pi +\alpha$.$$$$

Complete step by step answer:
We know that a trigonometric equation is an equation with trigonometric functions with unknown arguments as measure of angles. When we are asked to solve a trigonometric equation we have to find the all possible measures of unknown angles.$$$$
We know that the first solution of the trigonometric equation within the interval $\left[ 0,2\pi \right]$is called principal solution and using periodicity all possible solutions obtained with integer $n$ are called general solutions. The general solution of the trigonometric equation $\tan \theta =\tan \alpha$ with principal solution $\theta =\alpha$ are given by
$\theta =n\pi +\alpha$
We are given the following equation with tangent function with unknown angle constrained with the condition ${{0}^{\circ }}\le x\le {{360}^{\circ }}$ as
$3{{\tan }^{2}}x-1=0$
We divide both sides of above equation by 3 to have;

& \Rightarrow \dfrac{3{{\tan }^{2}}x-1}{3}=\dfrac{0}{3} \\\ & \Rightarrow {{\tan }^{2}}x-\dfrac{1}{3}=0 \\\ \end{aligned}$$ We use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for $a=\tan x,b=\dfrac{1}{\sqrt{3}}$ in the the above step and factorize the left hand side to have; $$\begin{aligned} & \Rightarrow \left( \tan x+\dfrac{1}{\sqrt{3}} \right)\left( \tan x-\dfrac{1}{\sqrt{3}} \right)=0 \\\ & \Rightarrow \tan x+\dfrac{1}{\sqrt{3}}=0\text{ or }\tan x-\dfrac{1}{\sqrt{3}}=0 \\\ & \Rightarrow \tan x=-\dfrac{1}{\sqrt{3}}\text{or }\tan x=\dfrac{1}{\sqrt{3}} \\\ \end{aligned}$$ We know from trigonometric table with basic values that the first $x$ in $\left[ 0,2\pi \right)$ for which $\tan x=\dfrac{1}{\sqrt{3}}$is $x=\dfrac{\pi }{6}$ and the first $x$ in $\left[ 0,2\pi \right)$ for which $\tan x=-\dfrac{1}{\sqrt{3}}$ is $x=\dfrac{5\pi }{6}$. So we have the principal solution for $\tan x=-\dfrac{1}{\sqrt{3}}$ is $x=\dfrac{5\pi }{6}$ and its general solution is $$\begin{aligned} & x=n\pi +\dfrac{5\pi }{6} \\\ & \Rightarrow x=n\times {{180}^{\circ }}+{{150}^{\circ }}........\left( 1 \right) \\\ \end{aligned}$$ Similarly the principal solution for $\tan x=\dfrac{1}{\sqrt{3}}$ is $x=\dfrac{\pi }{6}$ and its general solution is $$\begin{aligned} & x=n\pi +\dfrac{\pi }{6} \\\ & \Rightarrow x=n\times {{180}^{\circ }}+{{30}^{\circ }}........\left( 2 \right) \\\ \end{aligned}$$ We now choose an integer $n$ in a way that ${{0}^{\circ }}\le x\le {{360}^{\circ }}$. So for $n=0,1$ in equation (1) we have $x={{150}^{\circ }},{{330}^{\circ }}$ and for $n=0,1$ in equation (2) we have $x={{30}^{\circ }},{{210}^{\circ }}$. So the solutions of the given equation are $$\begin{aligned} & x={{30}^{\circ }},{{210}^{\circ }},{{150}^{\circ }},{{330}^{\circ }} \\\ & \Rightarrow x=\dfrac{\pi }{6},\dfrac{7\pi }{6},\dfrac{5\pi }{6},\dfrac{11\pi }{6} \\\ \end{aligned}$$ **Note:** We note that the trigonometric equation the unknown is always in radian variable unless otherwise mentioned like in this question. We note that $\tan x$ is not defined for $x=\left( 2n+1 \right)\dfrac{\pi }{2}$ and we would have discarded solutions $x=\left( 2n+1 \right)\dfrac{\pi }{2}$ if we would have obtained. The general solution of $\tan x=0$ is $x=n\pi $.