Is there a way to solve $\sin(x)=x$?

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.

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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 $x = 0$.

Hint: show that if $x\neq 0$ ($x$ real), $\left|\frac{\sin(x)}{x}\right|<1.$ I do not understand what you mean by "algebraically" so I will just leave this here and let you decide whether all solutions can be found "algebraically" or not.