SOLUTION: What values for q (0 £ q < 2p) satisfy the equation? 2sqrt2sinq + 2 = 0

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Question 627577: What values for q (0 £ q < 2p) satisfy the equation?
2sqrt2sinq + 2 = 0
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
2sqrt%282%29sin%28q%29+%2B+2+=+0
First we solve for sin(q). Subtracting 2:
2sqrt%282%29sin%28q%29+=+-2
Divide both sides by 2sqrt%282%29:
sin%28q%29+=+-1%2Fsqrt%282%29
Rationalizing the denominator:
sin%28q%29+=+%28-1%2Fsqrt%282%29%29%28sqrt%282%29%2Fsqrt%282%29%29
sin%28q%29+=+-sqrt%282%29%2F2
Now we solve for q. We should recognize -sqrt%282%29%2F2 as a special angle value for sin. We should recognize that the reference angle will be pi%2F4. And since our sin is negative and since sin is negative in the 3rd and fourth quadrants, we know that q must terminate there, too, with a reference angle of pi%2F4
So
q+=+pi+%2B+pi%2F4+=+4pi%2F4+%2B+pi%2F4+=+5pi%2F4 (for the third quadrant)
or
q+=+2pi+-+pi%2F4+=+8pi%2F4+-+pi%2F4+=+7pi%2F4 (for the fourth quadrant)