Supposedly there are four possible solutions. My professor provided a worksheet that says cos(2pi/3) and cos(4pi/3) both equal -1/2, and cos(pi/3) and cos(5pi/3) both equal 1/2. How would I go about finding the other two values that equal -1/2??

The trig functions have periods. That is, they repeat after so long.

Cosine has period of 2{\pi}

That is, they it repeats every 2Pi.

i.e. cos(2n\pi)=1 and so on

Use this to find the values where it equals cos(x)=-1/2.

x=\frac{2(3C+1){\pi}}{3}, \;\ and \;\ x=\frac{2(3C-1){\pi}}{3}

Now, enter integer values in for C. 0,1,2,3,.

Try C=0 and we get \frac{2{\pi}}{3}

Which cos(\frac{2\pi}{3})=\frac{-1}{2}

If C=1, we get \frac{8{\pi}}{3}

cos(\frac{8\pi}{3})=\frac{-1}{2}

And so on and so on.

Note the difference between \frac{8\pi}{3}-\frac{2\pi}{3}=2\pi

If you want the interval [0,2\pi], simply use the formula I gave you and root them out.

Does that help a little more?

I bet it does. All you have to do is plug in integer values of C to find what radian measure gives -1/2 for your cosine

Thank you. This is kind of helpful, but I am still pretty confused as to why there can only be 4 possible values... aren't there endless values?

That's because you only want those in a certain interval. Most likely from 0 to 2Pi

The two in that interval are \frac{2\pi}{3}, \;\ \frac{4\pi}{3}

For 1/2. they are \frac{\pi}{3}, \;\ \frac{5\pi}{3}

I am assuming 0 to 2Pi is the interval you are using. You must know that.

Look at the graph of cos(x). That will help visualize it.

I really appreciate all of your help galactus. I will find out what is going on with the wording of the question and let you know.

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