Question

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

22√sin q + 2 = 0

Answer 1

Answer:

`(3 \pi )/(4)` ,
`(7 \pi )/(4)`

Step-by-step explanation:
2√2 sin(q) + 2 = 0
2√2 sin(q) = -2
sin(q) =
`(-2)/(2 √(2) )`
sin(q) =
`(- √(2) )/(2)`

Now, we know that:
sin (45) =
`( √(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
`(3 \pi )/(4)`
or q = 270 + 45 = 315° which is equivalent to
`(7 \pi )/(4)`

Hope this helps :)

Answer 2

The value of q are
`(5\pi)/(4),(7\pi)/(4)`

C is correct.

Step-by-step explanation:

Given:
`2√(2)\sin q+2=0`

We need to solve for q.

`2√(2)\sin q+2=0`

Subtract 2 both side

`2√(2)\sin q=-2`

Divide both sides by
`2√(2)`

`\sin q=(-2)/(2√(2))`

`\sin q=(-1)/(√(2))`

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

`\sin q=-\sin (\pi)/(4)`

In III quadrant,
`q=\pi+(\pi)/(4)=(5\pi)/(4)`

In IV quadrant,
`q=2\pi-(\pi)/(4)=(7\pi)/(4)`

Thus, The value of q are
`(5\pi)/(4),(7\pi)/(4)`