# Suppose that two polar curves are given by: $R_1 = \cos(2\theta)$ and $R_2 = \sin(3\theta)$. Find the smallest positive solution exactly.

Suppose that two polar curves are given by: $R_1 = \cos(2\theta)$ and $R_2 = \sin(3\theta)$. Find the smallest positive solution exactly.

I know that we are looking for the smallest positive value where the two functions intersect. But I graphed it on a polar coordinate grapher, and I don't know how you would determine the correct solution.

Any help is appreciated!

Using formulas:

$$\sin{(3\theta)}=3\sin{(\theta)}-4\sin^3{(\theta)}$$ $$\cos{(2\theta)}=1-2\sin^2{(\theta)}$$

We can find a cubic equation for $\sin{(\theta)}$.

We have; \begin{aligned} \cos{(2\theta)}&=\sin{(3\theta)} \\ \cos{(2\theta)}-\sin{(3\theta)} &= 0 \\ 1-2\sin^2{(\theta)} - 3\sin{(\theta)}+4\sin^3{(\theta)} &=0 \end{aligned}

We can treat this as a cubic equation just with a variable $x$ for simplicity, so we need to solve, $$4x^3-2x^2-3x+1=0$$

This is found to have 3 distinct roots $$x_1=1, \qquad x_2=\frac{-(1+\sqrt{5})}{4}, \qquad x_3=\frac{\sqrt{5}-1}{4}.$$

Now we want $\theta$ to be the smallest positive value so we ignore $x_2$ and we can also see $0 < x_3 < x_1 \le1$. So we have $\sin(\theta)=\frac{\sqrt{5}-1}{4}$ and therefore $\theta = \frac{\pi}{10}$.

• Note, the last inequality holds only because $x_3 < x_1 \leq 1$. – Diego Robayo Dec 12 '14 at 2:41
• Thanks, forgot to mention that. Edited it accordingly. – Rammus Dec 12 '14 at 2:43

$$\cos2\theta=\sin3\theta=\cos\left(90^\circ-3\theta\right)$$

$$\implies2\theta=360^\circ m\pm(90^\circ-3\theta)$$ where $m$ is any integer

Taking the '+' sign, $$\implies2\theta=360^\circ m+(90^\circ-3\theta)\iff\theta=72^\circ m+18^\circ$$

Taking the '-' sign, $$\implies2\theta=360^\circ m-(90^\circ-3\theta)\iff-\theta=360^\circ m-90^\circ\iff\theta=-360^\circ m+90^\circ$$

Clearly, the smallest possible positive value of $\theta$ is $18^\circ$