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Plotting the equation $x^3-x^2 \sin(x)+\cos(x)$ I see that $x^3-x^2 \sin(x)+\cos(x)=0$ has only one real solution, is there a simpler way to see that it cannot have 3 real solutions?

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One consideration is the sin and cos are less than $1$ in magnitude, so the function may be compared to $x^3-x^2+1$ which also is close to $x^3 - x^2 = x^2(x-1)$. – adam W Oct 7 '12 at 13:11
up vote 1 down vote accepted

We have $f(x) = x^3-x^2\sin(x) +\cos(x)$, hence $f'(x)=(3-\cos(x))x^2-(2x-1)\sin(x)$, $f''(x)=(x^2-2)\sin(x) + 6x-(4x-1)\cos(x) $.

For $0\le x$ we have $\sin x\le x$, hence $f(x)\ge x^3-x^3+\cos(x)\ge \cos x$, hence any positive root of $f$ must be $\ge\frac\pi2$. But already for $x\ge\frac 3 2$, we have $f(x)\ge x^3-x^2-1=(x-1)x^2-1\ge(\frac32-1)\frac94-1=\frac18>0$. Therefore $f$ has no nonnegative root.

If $-\frac\pi2\le x\le 0$ then $\sin(x)\le \frac2\pi x$ and $\cos(x)\ge1+\frac2\pi x$, hence $f(x)\ge(1-\frac2\pi)x^3+1+\frac2\pi x$. The right hand side becomes $=0$ at $x=-1$ and has positive derivative. We conclude that $f(x)>0$ for $x> -1$.

Observe that $A\sin x+B \cos x=\sqrt{A^2+B^2}(\cos u\sin x+\sin u \cos x)=\sqrt{A^2+B^2}\sin(x+u)$ for some $u$, i.e. $$\tag1 |A\sin x + B\cos x|\le \sqrt {A^2+B^2}.$$ Therefore we see that $f(x)=x^3-x^2\sin x+\cos x$ can be estimated as $$ f(x)\le x^3+\sqrt{x^2+1}<x^3+\sqrt{x^2+1+\frac1{4x^2}}=x^3+\left|x+\frac1{2x}\right|,$$ i.e. for negative $x$ by $$ f(x)<x^3-x-\frac1{2x}=\frac{2x^2(x^2-1)-1}{2x} $$ The numerator is positive for $x^2>\frac{1+\sqrt 3}2$, i.e. for $x<-\sqrt{\frac{1+\sqrt 3}2}$, even more so for $x\le -\frac65$.

By the results so far, $f$ can have roots only in $(-\frac65,-1]$.

Between any two roots of $f$, there must be a root of $f'$. If $f$ has moe than one root, it must have at least three roots (counting multiplicity) because $f(x)\to\pm\infty$ as $x\to\pm\infty$, hence $f'$ must have two roots in $(-\frac65,-1]$ (again, counting multipliciites) and finally $f''$ must have at least one root in $(-\frac65,-1]$. Even a mild estimate for $\sin$ and $\cos $ in this interval should suffice to show $f''(x)<0$ here.

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Let $f(x) = x^3$ and $g(x) = x^2 \sin(x) - \cos(x)$. Clearly $$ |g(x)| \le x^2 + 1, $$ and hence a necesary condition for $f(x) - g(x) = 0$ is that $$ -\beta = \frac{1}{3}\big(-1 - \alpha_1^{-1/3} - \alpha_1^{1/3}\big) \le x \le \frac{1}{3}\big(1+\alpha_1^{1/3}+\alpha_2^{1/3}\big) = \beta $$ where $\alpha_1 = \frac{29-3\sqrt{93}}{2}$ and $\alpha_2 = \frac{29+3\sqrt{93}}{2}$.

Now, $$ g'(x) = x^2\cos(x) + (2 x + 1)\sin(x) $$ implies $g'(x) > 0$ from $0 < x \le \frac{\pi}{2}$, meaning that $g(x)$ is strictly increasing in $0 < x \le \beta < \frac{\pi}{2}$, achieving its maximum at $x = \beta$ and minimum at $x = 0$.

Given that $g(0) < f(0)$, $\,f$ striclty increasing, and $$ g(\beta) < \beta^2 \sin(\beta) < \beta^3 = f(\beta), $$ there is no positive solution to $f(x) - g(x) = 0$ for $x \in [0,\infty)$.

The case $ x < 0$ is a bit more interesting. First, we see that $$ g'(x) = 0 \quad \Longleftrightarrow \quad \frac{x^2}{2 x +1} = -\tan(x) $$ meaning that $g(x)$ has a (unique) critical point $x_c$ in $\big(-\beta,0\big)$; moreover, $x_c \in [-\frac{1}{2},0)$ and it's a maximum (it's easy to see that $g'(-\pi/4) > 0$ and $g'(-\pi/12) < 0$).

In $x \in [-\frac{1}{2},0)$, $g(0) < f(0)$ and $$ g(x) < -\cos(x) \le -\cos(-1/2) < -(1/2)^3 \le x^3 = f(x) $$ implying that there is no solution for $f(x) - g(x) = 0$ for $x \in [-\frac{1}{2},\infty)$.

Finally, $g'(x) > 0 $ for $x \in [-\beta, -\frac{1}{2}]$, and then there is one (and only one) solution in that interval.

Hope I didn't made any mistakes.

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It might be helpful to notice that it has an uneven number of solutions, since $x^3 \to \pm\infty$ for $x \to \pm\infty$. We can furthermore see that the function is not strictly even or odd (as it is composed of both an even function $\cos(x)$, and uneven functions $x^3, x^2\cdot\sin(x)$), this might hint at only one solution.

However, since the total function is made up of both polynomials of finite order and sinus/cosinus parts, it could have many more solutions - just compare it to the plot of

$$\frac{1}{100} x^3 - x^2\sin(x) + \cos(x).$$

So, to answer your question, there is no general way of deciding how many solutions such a function could have.

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