r=a sin(x)
r=a cos(2x)
How would I solve the following equation I know cos(2x)=2cos^2-1,1-sin^2(x)
But I am not sure where to begin.
Set them equal to each other $\displaystyle a sin(x) = a cos(2x) $, the a's cancel, so you get $\displaystyle sin(x) = cos(2x) $.
Now, $\displaystyle cos(2x) = 1 - 2*sin^2(x) $, so $\displaystyle sin(x) = 1-2 sin^2(x) $, so you got $\displaystyle sinx + 2 sin^2(x) = 1 $, so $\displaystyle sin(x)[1 + 2sin(x)] = 1 $. You can see that at $\displaystyle \frac{3\pi}{2} $ $\displaystyle sin(\frac{3\pi}{2}) = -1 $, so $\displaystyle sin(\frac{3\pi}{2})[1 + 2*sin(\frac{3\pi}{2})] = -1*(1 + 2(-1)) = -1*-1 = 1 $.
So your solution set would be $\displaystyle \frac{3\pi}{2} + 2\pi * k $ where K is an integer.
Let's not forget, you can NOT divide by 0. What should have been done is...
$\displaystyle \displaystyle \begin{align*} a\sin{(x)} &= a\cos{(2x)} \\ a\sin{(x)} - a\cos{(2x)} &= 0 \\ a\left[ \sin{(x)} - \cos{(2x)} \right] &= 0 \\ a = 0 \textrm{ or } \sin{(x)} - \cos{(2x)} &= 0 \end{align*}$
and then follow the abovementioned methods...