C2 Trig

Im currently on part b.
What does the question mean by saying q < 180?
Also, I plugged in the y/x values into the equations getting:

$\cos (100p - q)^{\circ} = 0$ for A(100, 0) and
$\cos (220p - q)^{\circ} = 0$ for B(220, 0).

$\therefore 100p - q = 90$ and $200p - q = 90$

When I try solving I get something like p = 0, which is rubbish. What am I doing incorrectly?
Thanks

PS just for the reference for part a, is the best approach to change the interval from 0 < x < 360 to 0 < 2x < 360, find the solutions, add 30 then divide by 2?

Didnt clock that mate, sorry

If I have

$\cos x = y$

$x=\cos^{-1} y$

AND

$x=360 - cos^{-1} y$

make sense?

$\cos(100p-q) = 0$

$\cos(220p-q) = 0$

You can work out the period of the function from this, normally roots of a cosine function would be $\pi = 180 ^\circ$ apart, here however they are only $120 ^\circ$ apart. So:

$p = \frac{180}{120} = \frac{3}{2}$

Now, the q is easily determined by noticing it's just going to translate the normal cosine function left/right (parallel to the x-axis) so work out the roots would normally lie with the function $\cos(\frac{3}{2}x)$, and you can work out the q from noting the coordinates of A, B and C.

P must be 0 you can see that this must be true by just looking at the simultaneous equations brah

p does not equal 0, it wouldn't be a cosine function if p = 0.

Oh I was looking at the simultaneous equation