Question

What are all the exact solutions of -3tan^2(x)+1=0? Give your answer in radians.

Answer 1

Answer: $x=\frac{\pi}{6}+n\pi,n\ \epsilon\ N$ and

$\ x=\frac{5\pi}{6}+n\pi,n\ \epsilon\ N$

Step-by-step explanation:

The given equation is $-3\tan^2(x)+1=0$

Subtract 1 on both sides, we get

$-3\tan^2(x)=-1$

Divide -3 on both sides, we get

$\tan^2(x)=\frac{1}{3}$

Taking square root on both sides, we get

$\tan\ x=\pm\frac{1}{\sqrt{3}}$

We know that $\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}$

Thus, when $\tan\ x=\frac{1}{3}\ then\ x=\frac{\pi}{6}+n\pi,n\ \epsilon\ N$

When  $\tan\ x=-\frac{1}{3}\ then\ x=\pi-\frac{\pi}{6}+n\pi=\frac{5\pi}{6}+n\pi,n\ \epsilon\ N$

Answer 2

   
$\displaystyle\\ -3\tan^2x + 1 = 0\\\\ -3\tan^2x = -1\\\\ \tan^2x = \frac{-1}{-3} \\\\ \tan^2x = \frac{1}{3} ~~~~~ \Big|~\sqrt{~~~} \\\\ \sqrt{\tan^2x} = \sqrt{\frac{1}{3}} \\\\ \tan x = \Big| \frac{1}{ \sqrt{3} } \Big|\\\\ \tan x = \Big| \frac{\sqrt{3}}{ 3 } \Big|\\\\ \tan x = \frac{\sqrt{3}}{ 3 }~~~\Longrightarrow~~x_1= \frac{\pi}{6} + k\pi,~~k \in N \\\\ \tan x = -\frac{\sqrt{3}}{ 3 }~~~\Longrightarrow~~x_2= \pi-\frac{\pi}{6} + k\pi = \frac{5\pi}{6} + k\pi,~~k \in N$