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?
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?
$f(x) = x^2-x\sin x-\cos x$ 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.
$f(0)=-1$ 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 $0$.
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.