Question

What are the values that satisfy the trigonometric equation for: 0≤q≤2π ? sin(∅)+tan(−∅)

Answer 1

Answer:

First choice: 0, pi

Step-by-step explanation:

I am assuming the equation is

$\sin q + \tan (-q) = 0$

(the =0 part is missing in the original problem)

$\sin q + \tan (-q) = 0\\\sin q + \frac{\sin (-q) }{\cos (-q)} = 0\\\sin q + \frac{-\sin q}{\cos q}=0\\\sin q = \frac{\sin q}{\cos q}\\\sin q \cdot \cos q = \sin q\,\,,\,\,\, 0\leq q \leq 2\pi\\\implies\\\cos q = 1 \implies q=\{0, 2\pi\}\\\mbox{or}\\\sin q = 0 \implies q=\{0, \pi, 2\pi\}\\\implies\\q=\{0,\pi,2\pi\}$

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)