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STEP Support Programme

STEP 2 2001 Q4 (spoilers!)

Hi,

I obtained 3 equations :
sin (2nx) = 0
cos(2nx) = 1
cos(2nx) = -1/2
When attempting the question I stated that m=n, and did not get a solution within the range. The integral solution treated these to be different which I don’t understand since they are both positive integers. Why is it also necessary to state each of the solutions have their own integer coefficients of pi (m, p and q). This meant I got no solutions in the range when I attempted the question once I had also used n = 2k and n = 2k +1.

Any help would be much appreciated!

Using the formulae for $\sin 2x=2\sin x \cos x$ and $\sin 3x = \sin x \cos 2x+ \sin 2x \cos x$ gives:
\[
\text{f}(x) = P \sin x + 2Q \sin x \cos x + R( \sin x \cos 2x+ \sin 2x \cos x)\\
=\sin x [P + 2Q \cos x + R(4\cos^2 x - 1)]
\]
The bit inside the square brackets is a quadratic in $\cos x$, and so you can show that there are no solutions to this part when $4Q^2-4(4R)(P-R)

For the next part, use a similar method to get:
\[
\text{g}(x) = \sin 2nx(2 + 2\cos 2nx - 4 \cos^2 2nx)
\]
Then the roots of $\text{g}(x)=0$ occur when $\sin 2nx = 0$ or when $2 \cos^2 2nx - \cos 2nx - 1=0$, i.e. when $\cos 2nx = 1$ or $-\frac 1 2$. Note that any solution of $\cos 2nx=1$ is also a solution of $\sin 2nx = 0$. This gives us the general solutions:
\[
2nx=m\pi \quad \text{or}\\
2nx = 2m\pi \pm \frac{2\pi} 3
\]
Note that here $n$ is really a fixed, unknown integer where as $m$ is a varying integer (there is a solutions when $m=1$, $m=2$ etc.)

Solving for $x$ gives:
\[
x=\frac{m\pi}{2n} \quad \text{or}\\
x = \frac {3m\pi \pm \pi} {3n}
\]
Considering the first set of solutions, if $x\lt \frac {\pi} 2$ then we need $\frac{m\pi}{2n} \lt \frac{\pi } 2$, which means that $m \lt n$ and so the largest possibly value of $m$ is $n-1$. The largest solution here is therefore $x=\frac {(n-1) \pi} {2n}=\frac {\pi} 2 - \frac {\pi}{2n}$.

For the other set of solutions, we have:
\[
\frac {3m\pi \pm \pi} {3n} \lt \frac{\pi} 2\\
\implies 6m \pm 2 \lt 3 n\\
\implies 6m \lt 3 n \mp 2
\]
Considering the solution $x = \frac {3m\pi - \pi} {3n}$ we have $m \lt \frac 1 2 n + \frac 1 3$. If $n$ is even (so $n=2k$ say) then we have $m \lt k + \frac 1 3$ and so $m=k=\frac 1 2 n$. This gives the solution $x=\frac{3\times \frac 1 2 n \pi - \pi}{3n}=\frac {\pi} 2 - \frac {\pi} {3n}$. This is bigger than the solution $x=\frac {\pi} 2 - \frac {\pi}{2n}$ from before.

If $n=2k+1$, then we have $m \lt k+ \frac 1 2 +\frac 1 3$ and so $m=\frac 1 2 (n-1)$. This gives the solution $x=\frac {\pi} 2-\frac {5\pi}{6n}$ which is smaller than the solution $x=\frac {\pi} 2 - \frac {\pi}{2n}$.

We now need to consider $x = \frac {3m\pi + \pi} {3n}$ which means we need $m

We therefore have a maximal value of:
\[
x=\frac {\pi} 2 - \frac {\pi} {3n} \quad \text{when }n \text{ is even}\\
x=\frac {\pi} 2 - \frac {\pi} {6n} \quad \text{when }n \text{ is odd}
\]