# Trig Help

1. Oct 30, 2009

### 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?

2. Oct 30, 2009

### 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

3. Oct 30, 2009

### abstrakt!

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