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I don't know what to do.

Would it be cot=1?

Well, that last equation simplifies to
\[2\cos^2x-2\cos x=0\]
What can you do with that?

Do you add 2 cos x to the other side?

Then divide by 2

2? Or is it cos?

Actually, it's both!

Factor out a 2cosx from both terms, and you get
\[2\cos x \;(\cos x - 1)=0\]
What's next?

Add 1 to the other side?

Oh so whhen they are in paranthesis, you can set them equal to 0?

The parentheses are there to show that the term (cosx - 1) is a factor of 0, as is 2cosx.

There's a big difference between
\[2\cos x\;(\cos x-1) \text{ and } 2\cos x \cos x -1\]