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Putting the equation $x^2 - x \sin(x) - \cos (x)$ into Wolfram Alpha, I am surprised that it has a nice parabolic shape. Also, it has two complex roots.

Question

Is it possible to tell, in a simple way, that it has no real roots?

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    Oh wow I totally got trolled by Wolfram Alpha! Thanks guys, I should have checked. -___-2012-09-17

3 Answers 3

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$f(x) = x^2-x\sin x-\cos x$ is even, $f(0) < 0$ and $\lim_{x \to \infty} f(x) = \infty$, hence the equation has at least two real roots. Also, $f'(x) = x(2-\cos x)$ satisfies $f'(x) \geq x \geq 0$ for $x \geq 0$, hence these are the only real roots.

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$f(0)=-1$ and $f(2)=4-2\sin(2)-\cos(2)\geq 4-2-1=1 \,.$

Then, by IVT it has a root between $0$ and $2$. Similarly, it has a second root between $-2$ and $0$.

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    Doh...I should have known to do this. :S Well, thanks!2012-09-17
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I think you are confused by treating this as a standard quadratic which can only have two real roots or a pair of conjugate complex roots. But this is not. As others commented it is more appropriate to use calculus to detect the distribution of roots on the real line. In the complex case your equation become $z^{2}-z\frac{e^{iz}-e^{-iz}}{2}-\frac{e^{iz}+e^{-iz}}{2}$ this equation is trancendatal and probably do not have easy solutions.

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    I see. Thanks for the comment! +1 :)2012-09-17