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 $$( r- \sin^2 2\theta)(r-\cos^2 2\theta)= 0.$$

Consider first the graph of

r - \sin^{2} 2 θ = 0

. For points on this graph, as

θ

increases from 0 to

\frac{π}{4}

we have

\sin^{2} 2 θ

increases from 0 to 1. Between

θ = \frac{π}{4}

and

\frac{π}{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

θ = \frac{π}{4}

Similarly the graph in the second quadrant is a 'petal' for $θ$ between $\frac{π}{2}$ and $π$ where $r$ takes positive values corresponding to $\sin^{2} θ$ . The graph in the third quadrant is a 'petal' for $θ$ between $π$ and $\frac{3 π}{2}$ and the graph in the fourth quadrant is a 'petal' for $θ$ between $\frac{3 π}{2}$ and $2 π$ .

Next consider first the graph of $r - \cos^{2} 2 θ = 0$ . This will be of the same form but rotated by $\frac{π}{4}$ corresponding to the phase shift between the graphs of $\sin 2 θ$ and $\cos 2 θ$ .

For points on this graph, as $θ$ increases from 0 to $\frac{π}{4}$ we have $\cos^{2} 2 θ$ decreases from 1 to 0. Between $θ = \frac{π}{4}$ and $\frac{π}{2}$ the value of $r$ increases from 0 to 1 and between $\frac{π}{2}$ and $\frac{3 π}{4}$ , 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 $θ$ between $\frac{3 π}{4}$ and $\frac{5 π}{4}$ . The next petal, symmetrical about the negative y-axis, is drawn for $θ$ between $\frac{5 π}{4}$ and $\frac{7 π}{4}$ and the remaining petal is completed for $θ$ between $\frac{7 π}{4}$ and $2 π$ .