Polar coordinates

/sci/, what would be the best method for solving the following type of problem, "Determine whether the point lies on the curve r^2 = sin(3x) [1,-5pi/6]"

i recommend killing yourself

>>2
Now you have two problems.

>>2

Op here, yeah I agree with ya. I should kill myself. Haha

Rewrite the equation in polar form, remembering that x = r*cos(theta):
r^2 = sin(3r*cos(theta))
Fill in the given coordinates:
r^2 = 1
sin(3*cos(-5pi/6)) = sin(3sqrt(3)/2) != 0

Conclusion, the point is not on the curve.

>>5
Correction: the last 0 should, of course, be 1

It's simple when the equation is simple and one can plug in the points and test for incidence but the book suggests testing for symmetry by converting to rectangular coordinates. The book states that the polar equation {r}^{2} = sin θ can be easily converted to \left ({x}^{2}+{y}^{2}\right )^{3} = {y}^{2}

I don't understand how that comes to be if r^2 = y^2 + x^2 and y=rsinθ

>>7
r^2 = \sin \theta \rightarrow r^3 = r \sin \theta \rightarrow \left (x^2 + y^2\right )^{\frac{3}{2}} = y \rightarrow \left (x^2 + y^2\right )^3 = y^2

>>8

Thaaaank you. :)