What are solutions to $$A \cos (\theta + \alpha) = \sin 2 \theta$$ where $\theta \in [0, 2\pi)$ and $A, \alpha$ are constants?
Graphically, this seems to have 2, 3, or 4 solutions, but I don't know how to find them analytically.
In fact, my goal isn't to find $\theta$, but rather characterize for which regions of $A, \alpha$ there are 4 solutions. I believe this will be one means of solving how many normals of a given ellipse intersect a given point.
Can we simplify via trig identities? $$q = A \cos \alpha \\ r = A \sin \alpha \\ q \cos \theta - r \sin \theta - 2 \sin \theta \cos \theta = 0$$
What about expanding the Taylor Series?
Comment: We may write:
$A \cos \theta \cos \alpha-\sin\theta \cos\theta=A \sin \theta \sin \alpha + \sin \theta \cos \theta$
$$\tan \theta=\frac {A \cos \alpha-\sin \theta}{A \sin \alpha - \cos \theta}\space\space\space\space\space\space\space\space\space\space(1)$$
Isolating $\theta$ leads to a mess of a trig. equation of degree 3. So we argue on relation (1):
1): $\theta=2k\pi\Longleftrightarrow A \cos \alpha=0$
Suppose $A\neq 0$, then $\cos \alpha=0\Rightarrow \alpha=(2k+1)\frac{\pi}2$; in $[0, 2\pi]; \alpha=\frac{\pi}2,.or. \frac{3\pi}2$
2): $\theta =\frac{\pi}4\Longleftrightarrow A \cos \alpha-\sin \theta=A \sin \alpha + \cos\theta\rightarrow A=\frac{\sqrt 2}{\cos \alpha-\sin \alpha}$
$\Rightarrow 0\leq\alpha <\frac {\pi}4$ or $\frac {\pi}4<\alpha\leq\frac{\pi}2$
(1) and (2) says there are real solutions for A and $\alpha$ in range $0<\theta<\frac {3\pi}2$
For $\theta=\frac{\pi}2$, relation (1) is undefined and there is no solutions for A and $\alpha$. That is equation has no solution like $(2k+1)\frac {\pi}2$ . These results are competent with the graphs shown below:
In this graph $A=\frac{\pi}4, B=\frac{\pi}2, C=\frac{3\pi}4, D=\pi\cdot\cdot\cdot$
Orange: $y=\sin 2\theta$
Blue: A=1, $\alpha=1$ which gives $y= \cos(\theta+1)$ $\theta\approx \frac {\pi}{16}$
Brown: A=2, $\alpha=1$ which gives $y+ \cos (\theta+1)$, $\theta\approx\frac{\pi}{10}$