How do you graph the system of polar equations to solve #r=2sintheta# and #r=2sin2theta#?

Question

1783 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-03T13:50:04

#( r, theta ) = ( 0, 0 )# and #( sqrt 3, pi/3 )#.

#r = 2 sin theta = 2 sin 2theta >= 0# gives

#sin theta - 2sin theta cos theta = sin theta ( 1 - 2 cos theta ) = 0#.

So, the factor # sin theta = 0# gives

#theta = kpi, k = 0, +-1, +-2, +-3, ...#.

Corresponding r = 0.

The factor # 2 cos theta - 1 = 0# sets

#cos theta = 1/2 = cos( pi/3)#, and this gives

#theta = 2npi +- pi/3, n = 0, +-1, +-2, +-3,...#

Here, only for #theta = 2npi + pi/3#,

#r = 2 sin theta = 2 sin 2theta = sqrt 3 > 0#.

Thus, combining both the sets, the common points are

#( r, theta ) = ( 0, 0 )# and # (sqrt 3, pi/3 )#. See graph.

Use #( x, y ) = r ( cos theta, sin theta )#.

graph{(x^2 + y^2 - 2y)((x^2 + y^2)^1.5-4xy)(x^2+y^2-.005)( (x-0.866)^2+(y-1.5)^2-.005)=0[-4 4 -2 2]}