Question

What values for q (0 < q < 2p) satisfy the equation?

2 square2 sin q + 2 = 0

Answer 1

\[a. \pi/4,3\pi/4
B.3\pi/4, 5\pi/4
C. \pi/4,7\pi/4 \]

Answer 2

\[2\sqrt{2}\sin(x)+2=0\]
\[2\sqrt{2}\sin(x)=-2\]
\[\sin(x)=-\frac{\sqrt{2}}{2}\]is a start

Answer 3

how would u finish that

Answer 4

i would think of an angle (or number) whose sine is \(-\frac{\sqrt{2}}{2}\)

Answer 5

Now pick angles in \((0,2\pi)\).