Question

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

Answer 1

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

Answer 2

Answer:
`x=(\pi)/(6)+n\pi,n\ \epsilon\ N` and

`\ x=(5\pi)/(6)+n\pi,n\ \epsilon\ N`

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)=(1)/(3)`

Taking square root on both sides, we get

`\tan\ x=\pm(1)/(√(3))`

We know that
`\tan(\pi)/(6)=(1)/(√(3))`

Thus, when
`\tan\ x=(1)/(3)\ then\ x=(\pi)/(6)+n\pi,n\ \epsilon\ N`

When
`\tan\ x=-(1)/(3)\ then\ x=\pi-(\pi)/(6)+n\pi=(5\pi)/(6)+n\pi,n\ \epsilon\ N`