Solve the equations, finding all $-2\pi \le \theta \le 2\pi$,

1. $${\tan }^2\theta -1=0$$
2. $$\sqrt{3}\tan \theta =2\cos \theta$$

Rearranging (a) and taking square roots gets the two possible cases

Sketch $y=\tan \theta$ (in radians) and use the graph to find all solutions.

For (b), recall that

Multiply both sides by $\cos \theta$.

Use ${\sin }^2\theta +{\cos }^2\theta =1$ to express ${\cos }^2\theta$ in terms of $\sin \theta$ - you end up with a quadratic to solve.