SOLUTION: solve for theta where 0<theta<2pi I) 2sin2theta= -1

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Question 823002: solve for theta where 0 I) 2sin2theta= -1
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Note: After seeing your thank you note I came back to my solution and checked. I'm sorry to say that I was wrong about the reference angle. A reference angle of pi%2F6 not pi%2F3 has a sin of 1/2. Everything else was correct. The problem below has been redone using the proper reference angle.

2sin%282theta%29+=+-1
First isolate the function and its argument by dividing both sides by 2:
sin%282theta%29+=+-1%2F2

We should recognize that 1/2 (positive or negative) is a special angle value for sin. Without our calculator we should know that a 1/2 for sin indicates a reference angle of pi%2F6. Since the 1/2 is negative and since sin is negative in the 3rd and 4th quadrants we should get the following general solution equations (for 2 theta):
2theta+=+pi%2Bpi%2F6%2B2pi%2An (for the 3rd quadrant)
2theta+=+-pi%2F6%2B2pi%2An (for the 4th quadrant)
The first equation will simplify:
2theta+=+7pi%2F6%2B2pi%2An
2theta+=+-pi%2F6%2B2pi%2An
Dividing both sides of both equations by 2:
theta+=+7pi%2F12%2Bpi%2An
theta+=+-pi%2F12%2Bpi%2An

Now we try various integer values for n as we search for specific solutions which are in the specified interval.
From theta+=+7pi%2F12%2Bpi%2An:
If n = 0 then theta+=+7pi%2F12
If n = 1 then theta+=+19pi%2F12
If n = 2 (or larger) then theta is too large for the interval
If n = -1 (or smaller) then theta is too small for the interval
From theta+=+-pi%2F12%2Bpi%2An:
If n = 0 (or smaller) then theta is too small for the interval
If n = 1 then theta+=+11pi%2F12
If n = 2 then theta+=+23pi%2F12
If n = 3 (or larger) then theta is too large for the interval

So the only solutions in the specified interval are: 7pi%2F12, 19pi%2F12, 11pi%2F12 and 23pi%2F12