$$\tan\theta + \sec\theta =2\cos \theta,\quad 0\le \theta\le 2\pi$$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.
Best Answer
Start with $$\tan(\theta) + \sec(\theta) = 2\cos(\theta).$$ Multiply both sides by $\cos(\theta)$ to get $$\sin(\theta) + 1 = 2\cos^2(\theta);$$ be warned that extraneous roots could be introduced where $\cos(\theta) = 0$, so you will hve to check these two roots separately.
Now use the Pythagorean identity to get
$$\sin(\theta) + 1 = 2 - 2\sin^2(\theta).$$ This is a quadratic-in-drag. Solve it; then check the two other places where $\cos(\theta) = 0$ separately. Beware of any domain considerations.