Let $\cos(bx)-\sin^2(x)-1=0$ has no zero point except 0. What's the value of b?
I have plotted many graphs of the function for several $b$. I think b can only be complex number. Is that right?
Substitute $\sin^2 x = 1-\cos^2 x$, and you get $\cos bx + \cos^2 x = 2$.
If $b$ is real, then $\cos bx=1$ and $\cos^2 x=1$. What does that tell you about $x$ and $bx$? What restriction on $b$ makes it impossible for any solutions other than $x=0$?