Posted by Anonymous on Thursday, November 22, 2007 at 9:01pm.
For the following equation, why cannot I not solve it? I keep getting an error.

2cosx  3 = 0

I first isolated for "x"
2cosx3 = 0
(2cosx/2) = (3/2)
x = cos^1(3/2)
x = error? why?

My textbook answer:
30 and 330 degrees


Math  Solving for Trig Equations  Reiny, Thursday, November 22, 2007 at 10:08pm
Make sure you copied the question correctly.
The way it stands
cos x = 3/2 has indeed no solution
(the sine and cosine function does not exceed a value >1 or < 1