How to solve $\tan^2 \theta − 2 \sec \theta = 2$ Solve the given equation. Find all solutions of the equation (express your answer in terms of k, where k is any integer).
$$\tan^2 \theta − 2 \sec \theta = 2$$
What do I do to solve for $\theta$?
update: I got
$cos\theta$ $= \frac {1}{3}$ and $cos\theta$ $-1$
and then:
$1.231+2\pi k,\space5.052+2\pi k,\space\pi +2\pi k$
Thank you!

Notice, we have $$\tan^2\theta-2\sec\theta=2$$ $$\sec^2\theta-1-2\sec\theta=2$$ $$\sec^2\theta-2\sec\theta-3=0$$ $$\sec^2\theta-3\sec\theta+\sec\theta-3=0$$
$$\underbrace{\sec^2\theta-3\sec\theta}\underbrace{+\sec\theta-3}=0$$ $$\sec\theta(\sec \theta-3)+(\sec\theta-3)=0$$ $$(\sec\theta-3)(\sec\theta+1)=0$$ $$\implies \sec\theta-3=0\iff \cos \theta=\frac{1}{3}\iff \color{blue}{\theta=2k\pi\pm\cos^{-1}\left(\frac{1}{3}\right)}$$
$$\implies \sec\theta+1=0\iff \cos \theta=-1\iff \color{blue}{\theta=2k\pi+\pi}$$
Where, $k$ is any integer & $\cos^{-1}\left(\frac{1}{3}\right)=1.231$
Your answers are obtained as follows,   $$\color{red}{\theta}=2k\pi+\cos^{-1}\left(\frac{1}{3}\right)=\color{red}{2k\pi+1.231}$$
$$\color{red}{\theta}=2k\pi+2\pi-\cos^{-1}\left(\frac{1}{3}\right)=\color{red}{2k\pi+5.052}\ \ \ \text{(since, }\ \cos\theta=\cos(2\pi-\theta))$$ & $$\color{red}{\theta=2k\pi+\pi}$$
