What values for q (0 ≤q≤2π)satisfy the equation?22√sin q + 2 = 0

Answer:
[tex] \frac{3 \pi }{4} [/tex] , [tex] \frac{7 \pi }{4} [/tex]

Explanation:
2√2 sin(q) + 2 = 0
2√2 sin(q) = -2
sin(q) = [tex] \frac{-2}{2 \sqrt{2} } [/tex]
sin(q) = [tex] \frac{- \sqrt{2} }{2} [/tex]

Now, we know that:
sin (45) = [tex] \frac{ \sqrt{2} }{2} [/tex]

From the ASTC rule, we know that the sine function is negative in the third and fourth quadrant.
This means that:
either q = 90 + 45 = 135° which is equivalent to [tex] \frac{3 \pi }{4} [/tex]
or q = 270 + 45 = 315° which is equivalent to [tex] \frac{7 \pi }{4} [/tex]

Hope this helps :)