Q:

For what $b$ does $\sin^2(x) - \cos(bx) + 1=0$ has only one solution?

For what $b$ does $f(x) = \sin^2(x) - \cos(bx) + 1=0$ has only one solution, if possible?
First, I noticed that $x=0$ is a root. If the equation has only one solution, then a sufficient condition would be that $f(x)$ is monotonic on $(-\infty,0)$ and $(0,\infty)$, which is not the case here. Also, I know that if $b=0$, then $f(x)$ has infinitely many solutions.
A plot seems to suggest that no such $b$ exists. But how should I prove it?

A:

You could use that $1-\cos(2a)=2\sin^2a$ so that the equation becomes
$$
\sin^2x+2\sin^2\frac{bx}2=0
$$
which means that you have to avoid that $x$ and $\frac b2x$ are simultaneously integer multiples of $\pi$. So chose $b$ as any irrational number.

