$$\tan\theta + \sec\theta =2\cos \theta,\quad 0\le \theta\le 2\pi$$Find all the possible solutions for the equations.
Multiply both sides by $\sec\theta - \tan \theta$. $$\implies (\tan\theta + \sec\theta)(\sec\theta - \tan\theta) = (\sec\theta -\tan\theta)2\cos \theta$$ $$\implies 1 = 2 -2\sin \theta$$ $$\implies \sin \theta=\frac12 \implies \theta = \arcsin\frac12$$Such a solution gets me two solutions $\frac{\pi}6$ and $\frac{5\pi}6$. But when I Wolfram it, I am supposed to get one more solution i.e $\frac{3\pi}2$, but at $\frac{3\pi}2$ $\tan \theta$ and $\sec\theta$ aren't defined.