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

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

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?

• Taylor series, maybe?
– user174622
Commented Jan 2, 2016 at 1:52
• I don't think you can "solve" this in the way you are used to doing with a finite number of algebraic operations (+ - * / ^). 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. Read up on transcendental functions on wikipedia if you are interested. Commented Jan 2, 2016 at 1:56
• I don't think you can solve it purely algebraically Commented Jan 2, 2016 at 1:59
• It is completely unclear what you mean 'algebraically solved'. If you can find all solutions exactly, what more do you want? Commented Jan 2, 2016 at 2:05
• Why does the only way 'it was done' have to be algebraic? In fact, do you think the solution $x=0$ is found "algebraically"? Hint: no. Commented Jan 2, 2016 at 2:08

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 $$x = 0$$.

• @SimpleArt Oh yeah, thanks. I added a bit to the end because I thought maybe that was the sort of process you were envisioning when you asked the question. Commented Jan 2, 2016 at 3:21
• Thanks for the addition. I wish there were a more algebraic way of doing it. Commented Jan 2, 2016 at 13:31
• I don't understand your argument that we can't solve this by algebraic means. Why does this argument fail when we try to solve $\sin x=0$ algebraically and we actually succeed? Commented Jan 3, 2016 at 14:11
• @Wojowu Can you explain or give an example of what you mean by your last sentence? This might be a simplification of Simple Art's question, but I think he was essentially asking for a way to evaluate $\sin(x) = x$ using only the techniques at the disposal of someone in early high school i.e. someone who hadn't learned calculus or analysis yet. Since there is no way to deduce through algebra that $\sin(x) = \sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{(2k+1)!}$, that example doesn't fit what he had in mind. Commented Jan 3, 2016 at 22:05

As you’re only considering real numbers, I think the easiest way to solve this is by splitting up cases and using inequalities in each case:

$$x=0$$ is clearly a solution, as $$\sin 0=0$$.

If $$x\in]0,1[$$, it follows from MVT that $$\exists c\in]0,1[: \cos c=\frac{\sin x-\sin 0}{x-0}$$.
As $$x,c\in]0,1[$$, it follows that $$1>\frac{\sin x}{x} \Leftrightarrow x>\sin x$$.

If $$x=1$$, then $$\sin 1 \neq 1$$.

If $$x>1$$, then obviously $$x>\sin x$$.

Now it is clear that if $$a$$ is a solution, then so is $$-a$$ (as $$\sin(-x) = -\sin x$$). Hence, there are no solutions with $$x<0$$

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.

• Thanks. More of a Calculus problem now? Commented Jan 2, 2016 at 2:18

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.

• Very smart looking Commented Aug 6, 2022 at 18:27