How many values of $\theta$ in the interval $0<\theta\le 2\pi$ satisfy\[1-3\sin\theta+5\cos3\theta = 0?\]$\textbf{(A) }2 \qquad$ $ \textbf{(B) }4 \qquad$ $ \textbf{(C) }5\qquad$ $ \textbf{(D) }6 \qquad$ $ \textbf{(E) }8$
Solution
Let $h(\theta)=5\cos3\theta+1-3\sin\theta$. The equation is $h(\theta)=0$, i.e.\ the graphs of $5\cos3\theta+1$ and $3\sin\theta$ meet.
The wave $5\cos3\theta+1$ has period $\frac{2\pi}3$ and oscillates between $-4$ and $6$; on $(0,2\pi]$ it has maxima at $\theta=0,\frac{2\pi}3,\frac{4\pi}3,2\pi$ and minima at $\theta=\frac{\pi}3,\pi,\frac{5\pi}3$.
At a maximum point $\cos3\theta=1$ so $h=6-3\sin\theta\ge3>0$. At a minimum point $\cos3\theta=-1$ so $h=-4-3\sin\theta\le-1<0$. Hence $h$ changes sign on each of the $6$ intervals between consecutive extreme points, so each contains at least one root.
Exactly one root in each interval: $h'(\theta)=-15\sin3\theta-3\cos\theta$. Where $|\sin3\theta|>\frac15$ we have $|15\sin3\theta|>3\ge|3\cos\theta|$, so $h'$ has the sign of $-\sin3\theta$ and $h$ is strictly monotone there.
Where $|\sin3\theta|\le\frac15$ we have $|\cos3\theta|\ge\frac{\sqrt{24}}5>0.97$. Near a maximum, $h\ge5(0.97)+1-3>2$, and near a minimum, $h\le-5(0.97)+1+3<0$, so $h$ has no zero in these parts.
So all roots lie where $h$ is strictly monotone, and each of the $6$ intervals contains exactly one root. (The endpoint $\theta=2\pi$ is not a root since $h(2\pi)=6$.) There are $6$ solutions, answer (D).
