1. ## Trig Equations

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

2. Originally Posted by greghunter
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 $\displaystyle \cos x$ in step 3.

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

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

Equating the two factors with zero gives two separate equations:

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

and

$\displaystyle 4\cos x+1=0$

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

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