Respuesta :

Louli
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 :)
isyllus

Answer:

The value of q are [tex] \frac{5\pi}{4},\frac{7\pi}{4}[/tex]

C is correct.

Step-by-step explanation:

Given: [tex]2\sqrt{2}\sin q+2=0[/tex]

We need to solve for q.

[tex]2\sqrt{2}\sin q+2=0[/tex]

Subtract 2 both side

[tex]2\sqrt{2}\sin q=-2[/tex]

Divide both sides by [tex]2\sqrt{2}[/tex]

[tex]\sin q=\frac{-2}{2\sqrt{2}}[/tex]

[tex]\sin q=\frac{-1}{\sqrt{2}}[/tex]

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

[tex]\sin q=-\sin \frac{\pi}{4}[/tex]

In III quadrant, [tex]q=\pi+\frac{\pi}{4}=\frac{5\pi}{4}[/tex]

In IV quadrant, [tex]q=2\pi-\frac{\pi}{4}=\frac{7\pi}{4}[/tex]

Thus, The value of q are [tex] \frac{5\pi}{4},\frac{7\pi}{4}[/tex]

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