Question

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

Answer 1

Answer:
$\frac{3 \pi }{4}$ , $\frac{7 \pi }{4}$

Explanation:
2√2 sin(q) + 2 = 0
2√2 sin(q) = -2
sin(q) = $\frac{-2}{2 \sqrt{2} }$
sin(q) = $\frac{- \sqrt{2} }{2}$

Now, we know that:
sin (45) = $\frac{ \sqrt{2} }{2}$

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 $\frac{3 \pi }{4}$
or q = 270 + 45 = 315° which is equivalent to $\frac{7 \pi }{4}$

Hope this helps :)

Answer 2

Answer:

The value of q are $\frac{5\pi}{4},\frac{7\pi}{4}$

C is correct.

Step-by-step explanation:

Given: $2\sqrt{2}\sin q+2=0$

We need to solve for q.

$2\sqrt{2}\sin q+2=0$

Subtract 2 both side

$2\sqrt{2}\sin q=-2$

Divide both sides by $2\sqrt{2}$

$\sin q=\frac{-2}{2\sqrt{2}}$

$\sin q=\frac{-1}{\sqrt{2}}$

Here, sin q is negative. Sine is negative in III and IV quadrant [0,2π ]

$\sin q=-\sin \frac{\pi}{4}$

In III quadrant, $q=\pi+\frac{\pi}{4}=\frac{5\pi}{4}$

In IV quadrant, $q=2\pi-\frac{\pi}{4}=\frac{7\pi}{4}$

Thus, The value of q are $\frac{5\pi}{4},\frac{7\pi}{4}$