Solving an iterated trig equation - Oxford Mathematics Admissions Test 2015

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[@RajSandhu-gm8iz]

Hi, what I did was let the stuff in brackets = y, so siny=0, hence y=0,pi,2pi etc. And -1

[@slytherinbrian]

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.

[@olivia242]

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.

[@mathoutloud]

You’re definitely correct, would be an unfortunate mistake to get caught on.

[@justabunga1]

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.