<s>[INST] 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!
 [/INST] 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{equation}
\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}
\end{equation}
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 $\theta = \frac{\pi}{10}$.
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