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I need to show that for every $ x>0 $:

$$ {3x - 2\sin(x)} \ne 0 $$

I have the intuition, and it's easy to show that for $ x\ge 1 $, since $ \sin $ is bounded then it holds, but I'm not sure how to show it for $ 0

It feels very simple on one hand but I still feel a bit stuck.

5 Answers 5

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Note that $3\cdot 0 - 2\sin(0) = 0$. Now, assume there is an $x_0 > 0$ such that $3x_0-2\sin(x_0) = 0$. What does the mean value theorem say?

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we have the known inequality $$\sin(x)\le x$$ from here we get $$2\sin(x)\le 2x<3x$$

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Lemma: If $0

Proof: enter image description here

Thus, if $02x>2\sin x$. Then $$3x-2\sin x>0$$

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    Whoever downvoted this answer draws disagreement from me. It's perfectly valid and avoids the need for calculus. – 2017-01-07
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    It doesn't avoid calculus because area of a sector cannot be defined without calculus. – 2017-01-07
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Let $f(x) = 3x, g(x) = 2\sin x$.

$f'(x) = 3, g'(x) = 2\cos x$.

$g'(0) = 2$

$g''(x) = -2\sin x$, so $g'(x)$ is decreasing for $0

Hence the straight line $y = 3x$ never intersects the curve $y = 2\sin x$ for positive $x$ in that range, and you already know how to show it for $x\geq 1$.

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Just thought of it , but i am not sure of it. Suppose $ 3x - 2\sin(x) = 0$ then we get $\frac{\sin(x)}{x} = \frac{3}{2}$ . Then when we take the limit $x$ tend to $0$ , then $\lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{3}{2}$ , thus we will be getting a contradiction.

Errors are welcome.