Answer:
First choice: 0, pi
Step-by-step explanation:
I am assuming the equation is
[tex]\sin q + \tan (-q) = 0[/tex]
(the =0 part is missing in the original problem)
[tex]\sin q + \tan (-q) = 0\\\sin q + \frac{\sin (-q) }{\cos (-q)} = 0\\\sin q + \frac{-\sin q}{\cos q}=0\\\sin q = \frac{\sin q}{\cos q}\\\sin q \cdot \cos q = \sin q\,\,,\,\,\, 0\leq q \leq 2\pi\\\implies\\\cos q = 1 \implies q=\{0, 2\pi\}\\\mbox{or}\\\sin q = 0 \implies q=\{0, \pi, 2\pi\}\\\implies\\q=\{0,\pi,2\pi\}[/tex]
So, the full solution for q is three values: 0, pi, 2pi. This matches the first choice (which is not complete, but that is ok)
