Convert Rectangular to polar form?

Question

(x^2+y^2)^2-9(x^2-y^2)=0

Answer 1

The relations between rectangular and polar coordinates are:
x = r·cos(φ)
y = r·sin(φ)

Therefore
(x² + y²)² - 9·(x² - y²) = 0
<=>
(r²·cos²(φ) + r²·sin²(φ))² - 9·(r²·cos²(φ) - r²·sin²(φ)) = 0
<=>
r⁴·(cos²(φ) + sin²(φ))² - 9·r²·(cos²(φ) - sin²(φ)) = 0
<=>
r⁴ - 9·r²·(cos²(φ) - sin²(φ)) = 0
(apply power reduction formula for sine and cosine)
<=>
r⁴ - 9 · r²·( (1 + cos(2φ))/2 - (1 - cos(2φ)/2) = 0
<=>
r⁴ - 9·r²·cos(2φ) = 0

Answer 2

Convert Rectangular to polar form.
(x² + y²)² - 9(x² - y²) = 0
__________

Remember the identities.

x² + y² = r²
x = rcosθ
y = rsinθ
___________

(x² + y²)² - 9(x² - y²) = 0

(r²)² - 9(r²cos²θ - r²sin²θ) = 0

r^4 - 9r²(cos²θ - sin²θ) = 0

r^4 - 9r²cos(2θ) = 0

Answer 3

(x^2+y^2)^2-9(x^2-y^2)=0

Let x = r cos t, y = r sin t .

[ r^2 (cos^2 (t) + sin^2 (t) )] ^2 - 9 r^2 [cos^2 (t) - sin^2 (t)] = 0

r^4 - 9r^2 [cos 2t] = 0 ,
where cos^2 (t) + sin^2 (t) =1 and cos^2 (t) - sin^2 (t) =cos(2t)