Non zero solution of $3x\cos(x) + (-3 + x^2)\sin(x)=0$

Question

How can I find exact non-zero solution of $3x\cos(x) + (-3 + x^2)\sin(x)=0$. Simple analysis and the below plot show that the equation has an infinite number of non-zero solutions.

Answer 1

There is an infinite number of solutions. Choose one region and find a solution there:

FindMinimum[Abs[3 x Cos[x] + (x^2 - 3) Sin[x]], {x, 9.5}] // Quiet

$\{\text{1.6217274456664654$\grave{ }$*${}^{\wedge}$-7},\{x\to 5.76346\}\}$

Answer 2

I am certain that there is no exact solution to this. However, by looking at regions where either $\sin(x)$ or $\cos(x)$ is small, I think that it can be shown that $f(x) =3x\cos(x) + (-3 + x^2)\sin(x) $ will change sign in these regions.

In particular, let $x = n\pi+y$ where $y$ is small.

Then $\sin(x) = \sin(y) \approx y$.

Also, $\cos(x) =\sqrt{1-\sin^2(x)} =\sqrt{1-\sin^2(y)} \approx\sqrt{1-y^2} \approx 1-\frac{y^2}{2} $.

Similarly, if $x = (n+\frac12)\pi+y$ where $y$ is small.

Then $\cos(x) = \sin(\frac{\pi}{2}-x) \approx y$ and $\sin(x) \approx 1-\frac{y^2}{2} $.

Try each of these in $f(x)$ and see which can be made small by making $y$ small.

Answer 3

There's probably no closed-form solution for the real roots you're looking for, although you have a chance with the complex solutions.

Let's notice that $3x \cos x$ and $\left( x^2 - 3 \right) \sin x$ have the same trigonometric argument. That means that the Harmonic Addition Theorem can be used: we can rewrite this as one trigonometric function.

HAT states that \begin{align}a \cos x + b \sin x &= A \cos \left(x + B \right), \end{align} where $a^2 + b^2 = A^2$ and $-\frac{b}a = \tan B$.

So, substituting $3x \cos x$ for $a$ and $x^2 - 3$ for $b$, we find that $A^2 = x^4 + 3x^2 + 9$ and $\frac{3 - x^2}{3x} = \tan B$.

This of course has infinitely many solutions, although the factor of $A$ only contributes some complex roots. Ignoring questions of signs and branch choice, let's declare that $A = \sqrt{x^4 + 3x^2 + 9}$ and $B = \tan^{-1} \frac{3 - x^2}{3x}$, or in other words $$3x \cos x + (x^2 - 3) \sin x = \sqrt{x^4 + 3x^2 + 9} \cos \left( x + \tan^{-1} \frac{3 - x^2}{3x} \right).$$

The problem can therefore be rewritten as finding the zeroes of $\sqrt{x^4 + 3x^2 + 9}$ and $\cos \left( x + \tan^{-1} \frac{3 - x^2}{3x} \right)$, although I doubt you'll find any good closed form for the latter.

This WolframAlpha plot looks like it agrees with your image.

Answer 4

Since we use Bessel J:

$$3x\cos(x)+(x^2-3)\sin(x)=0\iff-\frac{\sqrt\frac2\pi}{x^\frac52}(3x\cos(x)+(x^2-3)\sin(x))=\text J_\frac52(x)=0$$

The solution uses Bessel J Zero $\text j_{v,x}$:

$$3x\cos(x)+(x^2-3)\sin(x)=0 \iff x=\text j_{\frac52,\Bbb N}$$

where the $n$th natural number gives the $n$th solution shown here