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

 A: 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.
A: 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}$-7},\{x\to 5.76346\}\}$

A: 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.
A: 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
