Inverse of trig function

[Darya]

Hi guysl! I've been kinda struggling with finding inverses of trig functions on different intervals.
E.g. function [MATH]y=sin^2(x/2)[/MATH] on closed interval [MATH][2 \pi, 3 \pi][/MATH]. I start with taking square root on both sides.
[MATH]\sqrt{y}=\mid sin(x/2) \mid [/MATH] and it's [MATH]\sqrt{y}=- sin(x/2) [/MATH] because sin(x/2) is negative on that interval.
But then to get rid of sin I have to apply arcsin on both sides. But arcsin is defined on [-pi/2,pi/2]. I have no idea how I could shift this function and get inverse.
Please, any tips or ideas how to generally solve such exercises? I'd be thankful from the bottom of my heart.
Have a great day.

 

[Dr.Peterson]

One approach is to find all solutions for x/2 (giving two solutions, plus-or-minus 2n pi), and then restrict the interval as needed.

Another is to determine the interval in which x/2 must lie and use reference angles together with signs and supplements to find the angle in that interval.

Suppose you were just asked to solve sin^2(x/2) = -1/2 on the same interval; what would you do?

 

[Darya]

One approach is to find all solutions for x/2 (giving two solutions, plus-or-minus 2n pi), and then restrict the interval as needed.

Another is to determine the interval in which x/2 must lie and use reference angles together with signs and supplements to find the angle in that interval.

Suppose you were just asked to solve sin^2(x/2) = -1/2 on the same interval; what would you do?

Thanks! I understood the approach with determining the interval and using identities better. I guess sin^2(x/2) = -1/2 has no solutions in R though xd. But if I were asked that with 1/2, I'd first solve for x/2 and then picked the suitable interval among possible solutions. Many thanks!

 

[Dr.Peterson]

Thanks! I understood the approach with determining the interval and using identities better. I guess sin^2(x/2) = -1/2 has no solutions in R though xd. But if I were asked that with 1/2, I'd first solve for x/2 and then picked the suitable interval among possible solutions. Many thanks!

I was, of course, thinking sin(x/2) = -1/2, without the square, or perhaps sin(x/2) = -1/sqrt(2), because that made it more interesting ...