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.


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.
