It's been 20 years since I did trig, and this one seems a little tricky. How would I solve $$ \tan^2(x) -2\tan(x)=1 $$ with steps?
let $B = \tan(x)$
$\tan(x)$ varies with $x$, but ultimately is just a value
now rewriting the equation to give $B^2 -2B -1 = 0$, This will not factorise with integers, but solves to give $x = 1$ plus or minus square root $2$
from there, we use the $\arctan$ function and it tells you that (in radians) x = $\arctan(1 + \sqrt2)$ or $\arctan(1 - \sqrt2$) which will give you two roots every $2\pi$ radians, so you need to restrict the range of the function to get any real answers