How do you graph #r=5sin(3/4theta)#?

See
https://socratic.org/questions/how-do-you-create-a-graph-of-r-3-sin-7theta-5

#0 <= r <= 5#.

The period is #(8/3)pi#.. No graph for negative r., when

#(3/4)theta in ( pi, 2pi ) rArr theta in ((4/3)pi, (8/3)pi)#.

For four complete rotations, through #8pi#, three loop sought to

be created.

Upon setting #(8/3)pi = k(2pi)#, for least integer k, ,k = 3 #..

Use #sin 3theta = sin 4(3/4theta)# and expand both.

#3 cos^2theta sin theta - sin^3theta#

#= 4 (cos^3(3/4theta) sin(3/4theta)#

#- cos(3/4theta) sin^3(3/4theta)#

Now switch over to Cartesian form, using astutely

#sin(3/4theta) = r / 5, cos(3/4theta) = sqrt ( 1 - r^2/25),#

#r (cos theta, sin theta ) = ( x, y ) and r= sqrt ( x^2 + y^3 )#, to get

#3 x^2 y - y^3 = 4 ( x^2 + y^2 )^1.5(( 1 - (x^2 + y^2 )/25)^1.5#

#( (x^2 + y ^2)/25 )^0.5 -( 1 - (x^2 + y^2 )/25)^0.5#

#(( x^2 + y ^2 )/25)^1.5)#

I have paved the way. the Socratic graph is immediate.
graph{3x^2 y-y^3 - 0.0064 ( x^2 + y^2 )^2( 25 - (x^2 + y^2 ))^0.5( 25 - 2(x^2 + y^2 ) ) = 0[-10 10 -5 5]}
See graph of #r = cos (3/4)theta#:
graph{x^3-3xy^2-(x^2+y^2)^1.5(8(x^2+y^2)^2-8(x^2+y^2)+1)=0[-2 2 -1 1]}