For the following equation, why cannot I not solve it? I keep getting an error.
2cosx - 3 = 0
I first isolated for "x"
2cosx-3 = 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