Solving 8cosx-4=0: Find x-Values

[TonyC]

I am trying to identify the x-values that are solutions for the equation:
8cosx-4=0

I have come up with: x=pi/3,4pi/3
This doesn't seem correct but I am stumped.

 

[VietDao29]

Why 4pi/3?
Remember, if you need to solve:
\cos \alpha = \cos \beta = x
\Leftrightarrow \alpha = \pm \beta + k2\pi, k \in \mathbb{Z}
In other words,
\alpha = \pm \arccos x + k2\pi, k \in \mathbb{Z}
Viet Dao,

 

[TonyC]

so,
5pi/3

I C

Thank you

 

[HallsofIvy]

I am trying to identify the x-values that are solutions for the equation:
8cosx-4=0

I have come up with: x=pi/3,4pi/3
This doesn't seem correct but I am stumped.

8 cos x= 4 so cos x= 1/2. You could use a calculator but I think of half an equilateral triangle to observe that cos\left(\frac{\pi}{3}\right)= \frac{1}{2}. I then recall that "cos t" is the x coordinate of the unit circle parametrized by x= cos t, y= sin t. Drawing a vertical line at x= 1/2 I notice the vertical symmetry: cos\left(-\frac{\pi}{3}\right)= \frac{1}{2} also. That gives me two solutions, between -\pi and \pi[/tex]. If you want solutions between 0 and 2\pi, -\frac{\pi}{3}+ 2\pi= \frac{5\pi}{3} (NOT \frac{4\pi}{3}).<br /> <br /> Since cosine is periodic with period 2\pi, <b>all</b> solutions to 8cos x- 4= 0 are of the form \frac{\pi}{3}+ 2n\pi or \frac{5\pi}{3}+ 2n\pi for some integer n.