Solving the quadratic equation $$ r^2 - r + {\textstyle{1\over 4}} \sin^2 4\theta = 0$$ we get $$\eqalign{ r &= {-1 \pm \sqrt {1-\sin^2 4\theta}\over 2}\cr &= {1\pm \cos 4\theta \over 2} \cr &= \sin^2 2\theta,\quad \cos^2 2\theta .}$$ Hence the expression factorises to give $$( r- \sin^2 2\theta)(r-\cos^2 2\theta)= 0.$$
Polar graph



Consider first the graph of r- sin2 2q = 0. For points on this graph, as q increases from 0 to
p
4

we have sin2 2q increases from 0 to 1. Between
q = p
4

and
p
2

the value of r decreases from 1 to 0 so that the graph in the first quadrant is a 'petal' symmetrical about the line
q = p
4

.

Similarly the graph in the second quadrant is a 'petal' for q between
p
2

and p where r takes positive values corresponding to sin2 q. The graph in the third quadrant is a 'petal' for q between p and
3p
2

and the graph in the fourth quadrant is a 'petal' for q between
3p
2

and 2p.

Next consider first the graph of r- cos2 2q = 0. This will be of the same form but rotated by
p
4

corresponding to the phase shift between the graphs of sin2q and cos 2q.

For points on this graph, as q increases from 0 to
p
4

we have cos2 2q decreases from 1 to 0. Between
q = p
4

and
p
2

the value of r increases from 0 to 1 and between
p
2

and
3p
4

, as r decreases from 1 to 0, the 'petal' symmetrical about the y-axis is completed.

The next petal, symmetrical about the negative x-axis, is drawn for q between
3p
4

and
5p
4

. The next petal, symmetrical about the negative y-axis, is drawn for q between
5p
4

and
7p
4

and the remaining petal is completed for q between
7p
4

and 2p.