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

hb,uhliuhoiklStep-by-step explanation:...

What values for q (0 ≤q≤2π) satisfy the equation?

ptrlvn01

04.01.2022

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 $\frac{5\pi}{4},\frac{7\pi}{4}$

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