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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?

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up vote 3 down vote accepted


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 $x=0$?

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$b$ is not integer? – John Hass Nov 5 '12 at 15:15
Try reasoning it out, rather than just guessing. When is $\cos^2x =1$? When is $\cos bx = 1$? – Thomas Andrews Nov 5 '12 at 15:47

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