asd
It might be easier to switch to Cartesian coordinates. I'll explain soon - have to run and put a fire out right now.
Nope, false alarm. Same for the fire.
Any reason to think you have to get exact solutions to the equation?
Also, have you tried drawing the curves .... you realise that $\displaystyle r = 3 \sec (\theta - 30)$ is a line ...?
We have somewhat tired converting the equations to cartesian equations with conversion formulas, but it turned out..funny.
If you can do it, that would be wonderful.
We also know you can graph them as cartesian equations and get the intersections, but if we were to do that (according to our teacher) we have to have proof of why it works.
Thank you for your input, We'll look forward to more help.
My only other thought at this stage is to get an equation with r.
$\displaystyle r = 3 \sec (\theta - 30)$
$\displaystyle \Rightarrow \sqrt{3} r \cos \theta + r \sin \theta = 6$ .... (1).
$\displaystyle r = 1 + 5 \cos \theta$
$\displaystyle \Rightarrow \cos \theta = \frac{r - 1}{5}$ .... (2).
From (2), $\displaystyle \sin \theta = \pm \frac{\sqrt{25 - (r - 1)^2}}{5} = \pm \frac{\sqrt{(6 - r)(r + 4)}}{5}$ .... (3).
Substitute (2) and (3) into (1) and re-arrange into a quartic equation in r. You might get r from this (I'll cop that I haven't actually tried .....)
From the Pythagorean Identity: $\displaystyle \sin^2 \theta = 1 - \cos^2 \theta$
$\displaystyle 1 - \cos^2 \theta = 1 - \left( \frac{r - 1}{5} \right)^2 = 1 - \frac{r^2 - 2r + 1}{25}$
$\displaystyle = \frac{25 - r^2 + 2r - 1}{25} = \frac{-r^2 + 2r + 24}{25} = \frac{-(r^2 - 2r - 24)}{25}$ etc.
But on reflection I don't think you're gonna get anything even a little bit easy popping out of the resulting quartic equation in r .....
I'll keep it in the back of my mind.