First off, my practice problem did *not* have the 2 in it, which basically means that I didn’t think you’d need to use the chain rule to do it. Furthermore, you need to know about inverse trig functions for this, too. I really wasn’t worried about that even though we haven’t reviewed it yet, since it is pre-requisite material. So, I might consider giving everyone credit for this problem.

Having said all that, let’s throw caution to the wind and give it a go!

The tangent line is horizontal when f'(x)=0 and

f'(x) = 1+2\times5\cos(2x) = 1+10\cos(2x)

If we set that equal to zero and try to solve for x, we get

\cos(2x)=-1/10.

We might be tempted to apply the \arccos to both sides to get

2x = \arccos\left(-\frac{1}{10}\right) \: \text{ or } \: x = \frac{1}{2}\arccos\left(-\frac{1}{10}\right).

Let’s take a look at a graph, though:

Now, the red dot near \pi/4 is actually x=\arccos(-1/10)/2, that we found using the arccos. You actually want the green dot, though. It’s pretty clear from the symmetry that the blue dot is at -\arccos(-1/10)/2. Furthermore, the green dot is one period away from the blue dot and the period of \cos(2x) is \pi. So, I guess the answer is:

x = \pi - \arccos(-\pi/10)!