Trig Equations

greghunter · #1 $Old$ February 24th, 2009, 11:48 AM

Solve the following equations for 0<x<2pi

4cosx(cosx-1)=-5cosx

i managed to do-

4cos^2(x) - 4cosx = -5cosx
4cos^2(x) + cosx = 0
4cosx + 1 = 0
cosx = -1/4

x = 104.5

I used the cos graph but the answers I was given were-

x= 90, -90, 104, -104

bit confused

Reckoner · #2 $Old$ February 24th, 2009, 12:08 PM

Quote:

Originally Posted by greghunter $View Post$

i managed to do-

4cos^2(x) - 4cosx = -5cosx
4cos^2(x) + cosx = 0
4cosx + 1 = 0
cosx = -1/4

x = 104.5

I used the cos graph but the answers I was given were-

x= 90, -90, 104, -104

You have one of the answers, but you lost the other when you divided by $\cos x$ in step 3.

From $4\cos^2x + \cos x = 0$ we can factor

$\cos x(4\cos x+1)=0\text.$

Equating the two factors with zero gives two separate equations:

$\cos x=0$ (solution on $(0,2\pi)$ is $x=\pi/2$ )

and

$4\cos x+1=0$

$\Rightarrow\cos x=-\frac14$ (solution on $(0,2\pi)$ is $x=\arccos(-1/4)\approx1.8235\approx104.48^\circ$ ).

There should not be any negative solutions, since it was stated that $0<x<2\pi\text.$