What is the general solution of this trigonometric equation $$\tan^{2}\theta + \sec(2\theta)=1$$ from the following options:
a) $m\pi$
b) $n\pi\pm \frac{\pi}{3}$
c) $m\pi,n\pi\pm \frac{\pi}{3},\text{where }m,n\in \mathbb{Z}$
d) None of these
By examining the options in the given equation I get option a) b) c) satisfy the equation but I can't simplify it to get a general solution by myself.
My attempt:
$$\sec^{2}\theta -1+ \sec(2\theta)=1$$
Changing secant in terms of cosecant also doesn't help much instead make it complicated. I couldn't solve it further, thanks for help.
Best Answer
$$\tan^2t=1-\sec2t=1-\dfrac{1+\tan^2t}{1-\tan^2t}=-\dfrac{2\tan^2t}{1-\tan^2t}$$
$$\iff\tan^2t(3-\tan^2t)=0$$
Either $$\tan^2t=0\iff\tan t=0\iff t=m\pi$$
OR $$\tan^2t=3=\tan^2\dfrac\pi3$$
We know if $$\tan^2y=\tan^2A\iff y=r\pi\pm A$$ where $m,r$ are integers