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

Question

Note: Question was originally to solve it algebraically, though I've decided to change it to analytically due to the comments and answers.

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When trying to solve $\sin(x)=x$, the obvious first solution is $x=0$. There are, however, an infinite amount of complex values of $x$ we can try to find. However, we are going to ignore these.

I was wondering if there was a way to analytically solve for $x$ in $\sin(x)=x$. It does not appear to be possible, just like we can't solve $\cos(x)=x$ analytically or easily, but since $\sin(x)=x$ has such a simple exact answer, I wondered if there is a way you could do it.

So does there exist an analytic way we can solve this? If so, how? If not, how else would we solve it other than graphically?

Answer 1

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 $x = 0$.

Answer 2

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.

Answer 3

I see this is already answered, but I wanted to contribute with a very quick intuitive way of seeing this simply.

After the solution x=0, we just need to see that the slope of $\sin(x)$ which is $\cos(x)$ is $< 1$ in the regime that $x \le 1$ (after that is obvious there won't be solutions since $\sin(x)$ is bounded from -1 to 1, so any |x|>1 won't be solution) Check image for visuals of the slopes, in x $\epsilon$ [0,1]

So, no, there are not more real solutions apart from x=0.