We have this equation over real numbers: $$ x\sin x -1 = 0$$ First of all, find number of solutions in the interval $[-\pi, \pi]$. And then find the values of the solutions. (If it is possible, find the exact values.)
My try : I tried to use Euler's formula but it doesn't help.
Please Help!
Notice that $\sin(x)$ is a transcendental function. By nature, the inverse of the function $f(x)=x\sin(x)$ must then also be a transcendental function. Thus, the recommended path to finding the solution is a numerical method. One of the simplest of such is fixed-point iteration:
$$x\sin(x)=1$$
$$x=\csc(x)$$
$$x_{n+1}=\csc(x_n)$$
Let us assume that by graphing, the solution is something along the lines of $x\approx1.1$, or we would say $x_0=1.1$
$$\begin{align}x_1&=1.122073319\\x_2&=1.109875677\\x_3&=1.116515795\\x_4&=1.112870714\\\vdots\\x_{29}&=1.114157141\end{align}$$
$$x_{29}\sin(x_{29})=1.0000\dots$$
One could use the much better method known as Newton's method, which is basically taking tangent lines to approximate. Here, we would have
$$x_{n+1}=x_n-\frac{x_n\sin(x_n)-1}{\sin(x_n)+x_n\cos(x_n)}$$
Again, starting with $x_0=1.1$, we get
$$\begin{align}x_1&=1.114150789\\x_2&=1.114157141\end{align}$$
and thus we are already at the $29$th fixed-point iteration term. (lol?)
Upon the OP's request, I will show that the answer cannot be a rational multiple of $\pi$. This is simple, since it can be found that for $x$ being any rational multiple of $\pi$, $\sin(x)$ is algebraic. It is then found that $x\sin(x)$ will be transcendental. Clearly this cannot be the case, since $x\sin(x)=1$, an algebraic number.
Since $\sin(x)$ has no trivially known value in terms of special constants and such whenever $x$ is not a rational multiple of $\pi$, one cannot derive a closed form AFAIK.
This is not a full answer but it is too long for a comment.
As already said in answers and comments, such transcendental equations require numerical methods.
However, I cannot resist (once more) to show how good approximations can be obtained using the approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ which was proposed, more than $1400$ years ago, by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (see here).
Using it for the equation in the post, the problem is to find the solutions of $$16 x^3-4(4 \pi-1 ) x^2-4 \pi x+5 \pi ^2=0$$ and focuse on the positive roots (since, if $x$ is a solution, $-x$ is also a solution).
Using Cardano method for the solution of the cubic equation, we should get $$x_1=\pm 1.11495\qquad , \qquad x_2=\pm 2.77389$$ while the exact solutions, obtained using numerical methods would be $$x_1=\pm 1.11416\qquad , \qquad x_2=\pm 2.77260$$
Simpler, but less accurate, would be to approximate $\sin(x)$ by its Pade approximant built around $x=0$ $$\sin(x)=\frac{x-\frac{7 x^3}{60}}{1+\frac{x^2}{20}}$$ This will make the equation to be $$7 x^4-57 x^2+60=0$$ which is a quadratic in $x^2$ and the solutions would be $$x_1=\pm 1.11450\qquad , \qquad x_2=\pm 2.62693$$
