Can you show me how to find the rectangular equation of r=2costheta + 3 sin theta

All I need is an example to work off I was absent today from Pre-Calc and I have homework to do and no explanation with them-Please help with this one Thank you

what the heck is a rectangular equation?

hmm. you have given a polar coordinates equation, and want it y = f(x)?

cos theta = x/r
sin theta = y/r

r^2 = x^2 + y^2

r^2 = (2 cos t + 3 sin t)^2
= 4 cos^2 t + 12 sin t cos t + 9 sin^2 t
= 4 x^2/r^2 + 12xy/r^2 + 9y^2/r^2
r^4 = 4x^2 + 12xy + 9y^2

so, *x^2 + y^2)^2 = 4x^2 + 12xy + 9y^2

Not a pretty sight. The curve is cool, though

Certainly! I'd be happy to help you with that.

To find the rectangular equation of a polar equation, you can use the relationships between polar coordinates (r, θ) and rectangular coordinates (x, y), which are given by:

x = r * cos(θ)
y = r * sin(θ)

In your case, the polar equation is r = 2cos(θ) + 3sin(θ). To convert it to rectangular form, we'll substitute the values of x and y using the above relationships.

First, let's find x:

x = r * cos(θ)
x = (2cos(θ) + 3sin(θ)) * cos(θ)

Next, let's simplify it:

x = 2cos²(θ) + 3sin(θ) * cos(θ)

Now, let's find y:

y = r * sin(θ)
y = (2cos(θ) + 3sin(θ)) * sin(θ)

Again, let's simplify it:

y = 2sin(θ) * cos(θ) + 3sin²(θ)

Therefore, the rectangular equation is:

x = 2cos²(θ) + 3sin(θ) * cos(θ)
y = 2sin(θ) * cos(θ) + 3sin²(θ)

I hope this helps! Let me know if you have any more questions.