Problem
Consider the following geometric series:
$$G(x) = \cos(x) + 2\sin(x)\cos(x) + 4\sin^2(x)\cos(x) + \ldots$$
For which values of $b$ does the equation $G(x) = b$ have one or more solutions for $x \in [-\pi, \ \pi]$?
Thoughts
Since the series must converge, this imposes a restriction on $x$ to begin with. The ratio of the series is $2\sin(x)$, so the series converges when $-1 < 2\sin(x) < 1$, which means
$\displaystyle x\in \left[-\pi, -\frac{5\pi}{6}\right) \cup \left(-\frac{\pi}{6}, \frac\pi6\right) \cup \left( \frac{5\pi}{6}, \pi \right]$, within the same interval as the problem imposes.
We know that the sum of $G(x)$ would be $\frac{\cos(x)}{1-2\sin(x)} = b$, but this is where I hit a wall. I need to solve this equation, while also juggling the constraints on $x$ from the convergence ratios?
Since we have $-1 < 2\sin(x) < 1$, we also get $0 < 1-2\sin(x) < 2$, so it doesn't look like we run into any zero division problems either.
I'm not sure where to go from here.
Let $f(x)=\frac{\cos(x)}{1-2\sin(x)}$. For each $x$ in its domain, $$ f'(x)=\frac{\sin(x)^2}{(1-2\sin(x))^2}<0 $$ and therefore $f$ is strictly increasing on any interval.
Now, since $f(-\pi)=-1$ and $f\left(-\frac{5\pi}6\right)=-\frac{\sqrt3}4$, $\left[-1,-\frac{\sqrt3}4\right)$ is a subset of the range of $f$. Since $f\left(-\frac\pi6\right)=\frac{\sqrt3}4$ and $\lim_{x\to\frac\pi6^-}f(x)=\infty$, $\left(\frac{\sqrt3}4,\infty\right)$ is a subset of the range of $f$. Finally, since $\lim_{x\to\frac{5\pi}6^+}f(x)=-\infty$ and $f(\pi)=-1$, $(-\infty,-1]$ is a subset of the range of $f$. So, the equation $f(x)=b$ has a solution if and only if $|b|>\frac{\sqrt3}4$ and, for each such $b$, that equation has one and only one solution.