If the problem could be solved by purely algebraic means (with a finite number of steps), that would imply that $\sin(x)$ could be given a polynomial representation from which you could go about your usual routine of factoring to find the zeroes of the polynomial.
The interesting point here is that no such representation for $\sin(x)$ exists, unless you are okay with it being infinitely long.
The trigonometric functions like $\sin()$ and $\cos()$ are part of a category of transcendental functions--so called because they transcend the expressive power of algebra to describe them.
Here's a shot at solving it algebraically if we can cheat and use a result from calculus:
Given this identity:$$\sin(x) = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac{x^7}{7!} + \cdots $$
Subtract out your problem $\sin(x) = x$
$$0 = - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac{x^7}{7!} + \cdots $$
$$0 = x^3(- \frac {1}{3!} + \frac {x^2}{5!} - \frac{x^4}{7!} + \cdots) $$
$$x^3 = 0 \quad \mathrm{or} \quad (- \frac {1}{3!} + \frac {x^2}{5!} - \frac{x^4}{7!} + \cdots) = 0 $$
So now we have our "algebraic solution" that $x = 0$.
So, you tried to solve$$y+\frac1y=y^2+\frac1{y^2}.$$This is indeed equivalent to$$y^4-y^3+y-1=0.$$But\begin{align}y^4-y^3+y-1=0&\iff(y-1)(y^3+1)=0\\&\iff y=1(=e^0)\vee y=-1\left(=e^{\pi i}\right)\vee y=e^{i\pi/3}\vee y=e^{5\pi i/3}.\end{align}And so, yes, you do get the solutions $0$, $\pi$, $\frac\pi3$, and $\frac{5\pi}3$. Actually the whole set of solutions of the equation $\sin(\theta)=\sin(2\theta)$ is$$\pi\Bbb Z\cup\left\{\frac\pi3+2n\pi\,\middle|\,n\in\Bbb Z\right\}\cup\left\{\frac{5\pi}3+2n\pi\,\middle|\,n\in\Bbb Z\right\}.$$
Best Answer
I think the simplest numerical solution for such problems is Newton's method
You would look at the function
$f(x)=\sin(x)-x\ln(x)$
and now try to find the roots. One root is obvious $x=0$.
Newton's method works as follows. You need a start value $x_0$, which you guess, and then start a recursion.
$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$
To guess a starting point one could calculate some values of the function. Looking for a change in the sign.
For example $f(1)\approx 0.84$ and $f(2)\approx -0.47$
The intermediate value theorem guarantees a root in the intervall $(1,2)$, so one could take $x_0=1.5$ as the starting point.
Then
$x_1=1.5-\frac{\sin(1.5)-1.5\cdot\ln(1.5)}{\cos(1.5)-\ln(1.5)-1}\approx 1.7917$
$x_2\approx 1.7533$
$x_3\approx 1.7527$
$x_4\approx 1.7527$
It converges rather quickly.
With simple analytical arguments you can verify that these are the only two solutions. (There could be more of course)