Answer
$\theta=1.23+2k\pi$, $\theta=(2k+1)\pi$, $\theta=5.05+2k\pi$
Work Step by Step
$\tan^2 \theta-2\sec\theta=2$
$\sec^2 \theta-1-2\sec\theta=2$
$\sec^2 \theta-2\sec\theta-3=0$
$(\sec \theta-3)(\sec \theta+1)$=0
If $\sec \theta-3=0$, $\sec \theta=3$, so $\cos \theta=1/3$. Then $\theta=1.23+2k\pi$ or $\theta=5.05+2k\pi$.
If $\sec \theta+1=0$, then $\sec \theta=-1$, so $\cos \theta=\frac{1}{-1}=-1$. Then $\theta=\pi+2k\pi=(2k+1)\pi$.
