Question

How do you solve ${\tan ^2}x - 3 = 0$ and find all solutions in the interval $[0,2\pi )$ ?

Answer 1

In this question, we are given an equation containing one variable. It contains the tangent function and the tangent function is raised to the power 2 so it is a polynomial equation of degree 2 or we can also call it a quadratic equation. As the degree of this equation is 2, so tangent of x has exactly 2 values. We will take the constant term to the right-hand side and then square root both sides. This way we will get 2 values of $\tan x$ . Then using the knowledge of trigonometric ratios, we can find the value of x.

Complete step-by-step solution:
We are given that ${\tan ^2}x - 3 = 0$

$$\Rightarrow {\tan ^2}x = 3 \\\ \Rightarrow \tan x = \pm \sqrt 3 \\\$$

We know that $\tan \dfrac{\pi }{3} = \sqrt 3$ and $\tan \dfrac{{2\pi }}{3} = - \sqrt 3$ , so the general solution of $\tan x = \pm \sqrt 3$ is –
$x = \dfrac{\pi }{3} + \pi n,\,x = - \dfrac{\pi }{3} + \pi n$
When $n = 0$, we get $x = \dfrac{\pi }{3},\,x = - \dfrac{\pi }{3}$
When $n = 1$ , we get $x = \dfrac{{4\pi }}{3},\,x = \dfrac{{2\pi }}{3}$
When $n = 2$ , we get $x = \dfrac{{7\pi }}{3},\,x = \dfrac{{5\pi }}{3}$
We are given that x lies in the interval $[0,2\pi )$ so $x = - \dfrac{\pi }{3}$ and $x = \dfrac{{7\pi }}{3}$ is rejected.
Hence, when ${\tan ^2}x - 3 = 0$ , we get $x = \\{ \dfrac{\pi }{3},\,\dfrac{{2\pi }}{3},\,\dfrac{{4\pi }}{3},\,\dfrac{{7\pi }}{3}\\}$ .

Note: In the given question, we can find infinite values of x, but we have to find the values that lie in the interval $[0,2\pi )$ , that is, the values of x that are greater than or equal to 0 or smaller than $2\pi$ . In trigonometric functions, we are given an angle whose trigonometric value we have to find while in the inverse trigonometric function we are given the trigonometric value and we have to find the angle whose value it is. For example, $\tan x = \pm \sqrt 3$ is a trigonometric function and $x = {\tan ^{ - 1}}( \pm \sqrt 3 )$ is an inverse trigonometric function.