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 $x = 0$.
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)
Best Answer
There is no closed form in terms of any generally-accepted mathematical functions, as far as I am aware. In particular, the fixed point of the cosine function, i.e. the sole real solution of
$$\cos(x) = x$$
which is equivalent to
$$\sin\left(\frac{\pi}{2} - x\right) = x$$
is famously non-explicit. Hence, there can be no explicit formula for general values of $a$, $b$, $c$, and $d$ for the solutions of the equation you give.