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How many solutions does the equation $\cos^2x-\cos x-x=0$ have in the interval $\displaystyle \left[0,\frac\pi2\right]$?

Clearly, $x=0$ is a solution. Are there any other? I couldn't proceed after differentiating it to find the extremes.

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    $\begingroup$ Taking into account the answers you received, you could even prove that between -infinity and + infinity, x=0 is the only possible root. $\endgroup$ – Claude Leibovici Nov 29 '13 at 8:56
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You don't need to take the derivative for this.

We know that $x=0$ is an obvious solution and $\frac{\pi}{2}$ is not a solution. We also know that on the interval of $(0,\frac{\pi}{2})$, we have $0<\cos x<1$. Thus, we have the inequality $\cos^2 x<\cos x<\cos x+x$. Therefore, $\cos^2 x - \cos x-x<0$. This means that the only solution is $x=0$. Therefore, there is only $1$ solution.

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The derivative is $\sin x - \sin 2x - 1$ which is $\lt 0$ when $x \in [0,\frac{\pi}{2})$ (as $\sin 2x \gt 0$ and $\sin x \lt 1$)

You can fill in the rest.

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