.
Since we do not know what it looks lie, let's start with finding its key points. We can find #r# for values of #theta# equal to special angles and their multiples within the domain #0 < theta < 2pi#:
#r=costheta+sin2theta#
#theta=0, :. r=cos(0)+sin(0)=1+0=1#
#theta=pi/6, :. r=cos(pi/6)+sin(pi/3)=sqrt3/2+sqrt3/2=sqrt3=1.73#
#theta=pi/4, :. r=sqrt2/2+1=1.71#
#theta=pi/3, :. r=1/2+sqrt3/2=1.37#
#theta=pi/2, :. r=0+0=0#
#theta=(2pi)/3, :. r=-1/2-sqrt3/2=-1.37#
#theta=(3pi)/4, :. r=-sqrt2/2-1=-1.71#
#theta=(5pi)/6, :. r=-sqrt3/2-sqrt3/2=-sqrt3=-1.73#
#theta=pi, :. r=-1+0=-1#
Similarly, you can continue finding points on the grid by trying more values of #theta# until #2pi#.
Now, let's see what values of #theta# gives us #r=0#:
#costheta+sin2theta=0#
#costheta+2sinthetacostheta=0#
#costheta(1+2sintheta)=0#
#costheta=0, :. theta=pi/2, (3pi)/2#
#1+2sintheta=0, :. sintheta=-1/2, :. theta=(7pi)/6, (11pi)/6#
This means the graph will pass through the origin four times between #0 < theta < 2pi#
Now, let's see what the maximum and minimum values of #r# are by taking the derivative of the function, setting it equal to #0#, and finding the roots:
#(dr)/(d theta)=-sintheta+2cos2theta=0#
#2cos2theta=sintheta#
#2(1-2sin^2theta)=sintheta#
#4sin^2theta+sintheta-2=0#
Using the quadratic formula:
#sintheta=0.59307 and -0.84307#
#theta=0.635 and -1.003# radians
#theta~~pi/5 and -pi# are where the maximum and minimum #r# will occur which are #r~~1.76 and 0#
