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

Question 823002: solve for theta where 0 I) 2sin2theta= -1
Answer by jsmallt9(3758) (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 not has a sin of 1/2. Everything else was correct. The problem below has been redone using the proper reference angle.

First isolate the function and its argument by dividing both sides by 2:

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 . 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):
(for the 3rd quadrant)
(for the 4th quadrant)
The first equation will simplify:

Dividing both sides of both equations by 2:

Now we try various integer values for n as we search for specific solutions which are in the specified interval.
From :
If n = 0 then
If n = 1 then
If n = 2 (or larger) then is too large for the interval
If n = -1 (or smaller) then is too small for the interval
From :
If n = 0 (or smaller) then is too small for the interval
If n = 1 then
If n = 2 then
If n = 3 (or larger) then is too large for the interval

So the only solutions in the specified interval are: , , and
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