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Find all the points, if any, such that the curve $y=\sin(x-\sin x)$ has horizontal tangents at the $x$- axis.

My Solution

$$\frac{\mathrm{d} y}{\mathrm{d} x}\sin(x-\sin x)=(\cos (x-\sin x))(1-\cos x)$$

Let $\frac{\mathrm{d} y}{\mathrm{d} x}=0$ to find all the horizontal points.

We have, $(\cos (x-\sin x))(1-\cos x)=0$ which implies $(\cos (x-\sin x))=0$ or $(1-\cos x)=0$

Solving $(1-\cos x)=0$ first, we have $x= 2n\pi$ for all integers $n$.

Solving $(\cos (x-\sin x))=0$, I'm plain stuck. Any help thanks!

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up vote 2 down vote accepted

The question asks for the $x$-axis to be a (horizontal) tangent line of our curve.

So in addition to having $\frac{dy}{dx}=0$, we need to have $y=\sin(x-\sin x)=0$. We cannot have simultaneously the cosine and sine of $x-\sin x$ equal to $0$. This is a familiar fact about sine and cosine. They are never simultaneously $0$. For a proof, note that $\cos^2 t+\sin^2 t=1$.

So solutions of $\cos(x-\sin x)=0$ are no good for us. Good thing, too, since as you observed, finding the solutions looks like a hard problem!

So we need $1-\cos x=0$, that is, $x=2n\pi$ for some integer $n$. And we still need $y=\sin(x-\sin x)=0$. It should now not be hard to reach the right conclusion.

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Thanks! Didn't read the question carefully! – Yellow Skies Oct 3 '12 at 2:18

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