Hello

The question is solve on the interval $\displaystyle -\pi\leq x \leq\pi$

$\displaystyle \sec^2(x) = 4\tan(x)$

using the identity $\displaystyle \sec^2(x) = \tan^2(x) +1$

I get:

$\displaystyle \tan^2(x) - 4\tan(x) +1 = 0$

I can't factor that, so I completed the square and got.

$\displaystyle \tan(x) = 2\pm\sqrt{3}$

It looks like that will give me the solutions, but I've never completed the square on a trigonometric function before and just wanted to check that it's the right way to proceed.

Thank you.