Math  Solving for Trig Equations
posted by Anonymous .
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


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
Respond to this Question
Similar Questions

math
tanx+secx=2cosx (sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 = 2(1sin^2x) 2sin^2x + sinx1=0 (2sinx+1)(sinx1)=0 x=30 x=270 but if i plug 270 back into the original equation … 
Trig
cos2x(2cosx+1)=0 Solve for x. 
trig
prove that (2cosx+1)(2cosx1)= 1 +2cos2x 
trig
how do you graph the equation y=2cosx? 
Calculus
Determine the concavity and find the points of inflection. y=2cosx + sin2x 0≤x≤2pi y'=2sinx + 2cos2x y"=2cosx4sinx How do I find the IP(s)? 
Trig
Solve for x in the interval [0,2pi) sin^2x+2cosx=2 
Mathj C30
Solving Equations Containing Circular Functions : Solve 2cosx sinx + sinx = 0 when x is between 0 and 2pi 
Math
Find the limit of (2cosx+3cosx2)/(2cosx1)as x tends to pi/3. 
Pre Calculus
Having trouble verifying this identity... cscxtanx + secx= 2cosx This is what I've been trying (1/sinx)(sinx/cosx) + secx = 2cosx (1/cosx)+(1/cosx) = 2 cosx 2 secx = 2cosx that's what i ended up with, but i know it's now right. did … 
derivatives
x^2cosx2xsinx2cosx