Is it possible to solve $x\sin(x) = 7$ algebraically? 
$$x\sin(x)=7$$

The only method to solve this that comes to my mind is to draw the left side of this equation. However, there is an infinite number of solution. Is it possible to come up with something like $x=x_o+nT$ where T is the "period"?
 A: As said in comments, only numerical methods will allow to find the almost exact solutions.
However, the solutions can be approximated if, as you surely noticed from the plot of function $$f(x)=x\sin(x)-7$$ the roots are very close to $n\pi$ as soon as $n\gt 2$ (we just focus on the positive roots since if $x$ is a root, $-x$ is a root too). 
Expanding $f(x)$ as a Taylor series around $x=n\pi$, we get $$f(x)=(\pi  n \sin (\pi  n)-7)+(x-\pi  n) (\sin (\pi  n)+\pi  n \cos (\pi  n))+(x-\pi 
   n)^2 \left(\cos (\pi  n)-\frac{1}{2} \pi  n \sin (\pi  n)\right)+O\left((x-\pi 
   n)^3\right)$$ Using $\sin(n\pi)=0$, $\cos(n\pi)=(-1)^n$ and ignoring the higher order terms, we then can solve the quadratic for the different cases where $n$ is odd or even. Doing it, for even values of $n$ the equation is $$F_{2n}=x^2-2 n \pi   x-7=0 \implies x_{2n}=n\pi  +\sqrt{\pi ^2 n^2+7}$$ 
$$F_{2n+1}=x^2-(2n+1) \pi   x+7=0 \implies x_{2n+1}=\frac 12 \left((2n+1)\pi +\sqrt{(2n+1)^2 \pi^2-28}\right) $$
For illustration purposes $$x_{10}=5 \pi +\sqrt{7+25 \pi ^2}\approx 31.6372$$ $$x_{11}=\frac{1}{2} \left(11 \pi +\sqrt{121 \pi ^2-28}\right)\approx 34.3538$$ while the "exact" solutions would be $31.6390$ and $34.3523$.
You could have other approximations making one single iteration of Newton method: this would give as estimates $$x_{2n}=2 n\pi  +\frac{7}{2n \pi  }\qquad x_{2n+1}=(2 n+1)\pi  -\frac{7}{(2n+1) \pi  } $$ which would give $x_{10}\approx 31.6387$ and $x_{11}\approx 34.3550$.
