# How many distinct real root does the equation $x^{4}-x^{3}\cdot\sin(x)-x^{2}\cdot\cos(x)=0$ have?

How many distinct real root does the equation $x^{4}-x^{3} \cdot \sin(x)-x^{2} \cdot \cos(x)=0$ have?

Is there any quick solution(less than 2 minutes)?

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From looking at the clock: Apparently not :) – Hagen von Eitzen Oct 22 '12 at 15:47

You can factor out $x^2$ to get $x^2(x^2-x \sin x - \cos x ) = 0$. So, clearly, there is a root of multiplicity $2$ at $x=0$. Letting $f(x) = x^2-x \sin x - \cos x$, we note that $f$ is even, $f(0) <0$, $\lim_{x\to\infty} f(x) = \infty$ and $f'(x)> 0$ when $x>0$. Consequently $f$ has one real root in $(0,\infty)$.

So the answer is 3 distinct real roots.

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Nice Solution copper.hat. – juantheron May 5 '15 at 3:49

First observe that your equation can be rewritten as: $$x^2\cdot f(x), \qquad f(x):= x^2 -x\sin(x) -\cos(x).$$

In specific it has the root $x=0$ with multiplicity $2$ and the roots of $f(x)$. To understand how many real roots has $f(x)$ let us compute its derivative:

$$\partial_x f(x)= x(2-\cos(x)).$$

Then the derivative is zero only for $x=0$. Since $\lim_{x \rightarrow \pm \infty} f(x) = +\infty$ we deduce that $f(x)$ will have 2 distinct real roots if $f(0)<0$, a unique real root with multiplicity $1$ if $f(0)=0$ and no real roots if $f(0)>0$. (Can you justify this statement?)

In our case $f(0)= -cos(0)=-1<0$. So $f(x)$ has two distinct real roots and both of them are distinct from zero. (Again, can you justify this last statement?) In specific your original function has 3 distinct real roots.

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