Trig Help

[abstrakt!]

1. The problem statement, all variables and given/known data

4 \ sin \ \theta \ = \ 3 \ csc\ \theta

3. The attempt at a solution

sin\ \theta \ = \ \frac {3}{4} \ csc \ \theta

sin^2 \ \theta \ = \ \frac {3}{4}

sin \ \theta \ = \ \pm \ \frac {\sqrt{3}}{2}

30 \ \deg \ in \ QI, \ 150 \ \deg \ in \ QII, \ 210 \ \deg \ in \ QIII, \ 330 \ \deg \ in \ QIV

Am I doing this correctly?

[lurflurf]

Almost, but I think you have confused sine and cosine
sin(30 degrees)=1/2
sin(60 degrees)=sqrt(3)/2
cos(30 degrees)=sqrt(3)/2
cos(60 degrees)=1/2

[abstrakt!]

Almost, but I think you have confused sine and cosine
sin(30 degrees)=1/2
sin(60 degrees)=sqrt(3)/2
cos(30 degrees)=sqrt(3)/2
cos(60 degrees)=1/2

I understand the difference between the two, I must have hit the wrong button in my calculator. Thanks brotha.