I just drew the graph of cos with two periods between 0 and 2 pi, and streched and shifted up to be between 0 and 4, and counted the number of times the graph crossed pi (4 times) or touched the x axis (twice), since those are the only values where sin of that is zero.
The main thing to watch out for with this type of question is the limits of x because it's inclusive on both boundaries which in some cases results in an extra answer but that isn't the case here.
Before you try to include the answers that is within the interval, take the equation and solve for x as if you were to find all of the solutions. Here, we know that 2cos(2x)+2=nπ. Now, 2cos(2x)=nπ-2, so cos(2x)=(nπ-2)/2. Convert cos(2x) to either of the two forms. I'll go with 2(cos(x))^2-1=(nπ-2)/2. Add to 1 both sides and then divide by 2. What we have there is (cos(x))^2=nπ/4, so cos(x)=±√(nπ)/2, so x=2kπ±arccos(±√(nπ)/2) where k and n are integers. Looking at the interval there, if you plug in some integer values for both k and n, we only see that there are 2 solutions within that interval, which is choice A.
