Convert r=1+3sin2x from a polar equation to a rectangular equation.

I assume you mean

r = 1 + 3 sin 2t
where t = theta

x = r cos t
y = r sin t

sin 2 t = 2 sin t cos t
so
r = 1 + 6 sin t cos t
r = 1 + 6 (y/r)(x/r)
r^3 = 1 + 6 x y
but r^2 = x^2+y^2
r= (x^2+y^2)^.5
so
(x^2+y^2)^(3/2) = 1 + 6 x y
(x^2+y^2)^3 = 1 + 12 x y + 36 x^2 y^2
you can multiply the left out

wouldnt the "1"in the 1+.... in the 4th step become r^2 when you multiply both sides by r^2.

You are right

r^3 = r^2 + 6 x y
(x^2+y^2)^3 = (x^2+y^2)^2 + 6 x y

x^6 + 3 x^4y^2 + 3 x^2 y^4 + y^6

= x^4 + 2 x^2y^2 + y^4 + 6 x y

Thanks a lot for your help

To convert the given polar equation r = 1 + 3sin(2x) to a rectangular equation, we can use the following trigonometric identities:

x = r*cos(theta)
y = r*sin(theta)

First, let's solve for r by multiplying both sides of the equation by r:

r^2 = r(1 + 3sin(2x))

Next, we can substitute x with r*cos(theta) and simplify the equation:

r^2 = r(1 + 3sin(2r*cos(theta)))

Expanding the equation further:

r^2 = r + 3rsin(2r*cos(theta))

Simplifying the equation by dividing both sides by r:

r = 1 + 3sin(2r*cos(theta))

Now, let's substitute r with sqrt(x^2 + y^2) and simplify the equation:

sqrt(x^2 + y^2) = 1 + 3sin(2sqrt(x^2 + y^2)*cos(theta))

To convert sin(2sqrt(x^2 + y^2)*cos(theta)) to rectangular form, we'll use the double angle identities:

sin(2theta) = 2sin(theta)*cos(theta)

Substituting back into the equation:

sqrt(x^2 + y^2) = 1 + 3(2sin(sqrt(x^2 + y^2)*cos(theta))*cos(sqrt(x^2 + y^2)*cos(theta)))

Simplifying further:

sqrt(x^2 + y^2) = 1 + 6sin(sqrt(x^2 + y^2)*cos(theta))*cos(sqrt(x^2 + y^2)*cos(theta))

To eliminate sqrt(x^2 + y^2) from the equation, we can square both sides:

x^2 + y^2 = (1 + 6sin(sqrt(x^2 + y^2)*cos(theta))*cos(sqrt(x^2 + y^2)*cos(theta)))^2

Expanding and simplifying:

x^2 + y^2 = 1 + 12sin(sqrt(x^2 + y^2)*cos(theta))*cos(sqrt(x^2 + y^2)*cos(theta)) + 36sin^2(sqrt(x^2 + y^2)*cos(theta))*cos^2(sqrt(x^2 + y^2)*cos(theta))

Finally, this is the rectangular equation that represents the polar equation r = 1 + 3sin(2x):

x^2 + y^2 = 1 + 12sin(sqrt(x^2 + y^2)*cos(theta))*cos(sqrt(x^2 + y^2)*cos(theta)) + 36sin^2(sqrt(x^2 + y^2)*cos(theta))*cos^2(sqrt(x^2 + y^2)*cos(theta))

Note that this equation is not easily simplified since it involves trigonometric functions and variables in both the x and y terms.