Obviously you can draw a graph, but how would you prove this with calculus?
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2Differentiate $3x-2-\sin x$. See that the derivative is always positive. Having two different roots would violate the Mean Value Theorem. – hmakholm left over Monica Nov 21 '14 at 19:29
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When is the RHS inside the image $\sin(\mathbb R)$? How does $\sin$ behave there? EDIT: Henning's answer is more straight-forward. – Nov 21 '14 at 19:30
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1And having no roots at all would violate the Intermediate Value Theorem. – Marc van Leeuwen Nov 21 '14 at 19:34
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Since $|\sin'(x)| \le 1$ you have $|\sin x_1 - \sin x_2| \le |x_1 -x_2|$.
Suppose $x_1, x_2$ satisfy the given equation. Then $|\sin x_1 - \sin x_2| = 3 |x_1-x_2|$ and the only way $3 |x_1-x_2| \le |x_1 -x_2|$ can be satisfied is if $x_1 = x_2$.
As an aside, if we let $\phi(x) = {1 \over 3} (\sin x +2)$, we see that $x$ solves the equation iff $\phi(x) = x$. Note that $|\phi'(x)| \le {1 \over 3}$, hence $\phi$ is a contraction map and has a unique solution which can be found by iterating $x_{n+1} = \phi(x_n)$.
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