<s>[INST] Is there a way to solve the equation $\sin x = x\ln x$ analytically? 
Is there a way to solve the equation $\sin x = x\ln x$ numerically or analytically?

The only way I have been able to solve this is using a graphic calculator like Desmos, but is there another way to solve this?
 [/INST] This is a transcendental equation; this means no hope for a closed form solution and numerical methods are reaquired. So, considering that you looks for the zero's of function $$f=\sin (x)- x\log(x)$$ for which $x=0$ is a trivial solution. By inspection, the solution is between $1$ and $2$. Being lazy (myself), make a Taylor expansion around $x=\frac \pi 2$; this would give
$$f=\left(1-\frac{1}{2} \pi  \log \left(\frac{\pi }{2}\right)\right)+\left(x-\frac{\pi
   }{2}\right) \left(\log \left(\frac{2}{\pi }\right)-1\right)-\frac{(2+\pi )
   \left(x-\frac{\pi }{2}\right)^2}{2 \pi }+\frac{2 \left(x-\frac{\pi
   }{2}\right)^3}{3 \pi ^2}+\left(\frac{1}{24}-\frac{2}{3 \pi ^3}\right)
   \left(x-\frac{\pi }{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^5\right)$$ Now, using series reversion
$$x=\frac{\pi }{2}-\frac{f-1+\frac{1}{2} \pi  \log \left(\frac{\pi }{2}\right)}{1+\log
   \left(\frac{\pi }{2}\right)}-\frac{(2+\pi ) \left(f-1+\frac{1}{2} \pi  \log
   \left(\frac{\pi }{2}\right)\right)^2}{2 \pi  \left(1+\log \left(\frac{\pi
   }{2}\right)\right)^3 }+O\left(\left(f-1+\frac{1}{2} \pi  \log \left(\frac{\pi
   }{2}\right)\right)^3\right)$$
Making $f=0$ leads to $x=1.74843$ while the "exact" solution (obtained by Newton method) is $1.75268$
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