SOLUTION: I've had some difficulties with these interval problems. I can't seem to come up with all of the correct answers. Usually each question has four answers but I can't seem to find th

Question 962972: I've had some difficulties with these interval problems. I can't seem to come up with all of the correct answers. Usually each question has four answers but I can't seem to find them.
Could someone please help me with the following question:
Find all solutions in the interval [0,2π): 4√2*sin(x/2)= -4
Enter the solutions in radians, or enter the null set.
Thank you in advanced.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find all solutions in the interval [0,2π): 4√2*sin(x/2)= -4
Enter the solutions in radians, or enter the null set.
-------
sin(x/2) = -4/(4sqrt(2)) = -1/sqrt(2)
-----------
For that sin value the angle must be in QIII or QIV
and must have a reference angle of pi/4.
------
In QIII you get x/2 = (5/4)pi
Then x = (5/2)pi
----
In QIV you get x = (7/4)pi
Then x = (7/2)pi
----------------------------
Cheers,
Stan H.

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