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?
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$\begingroup$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 $x = 0$.
$\endgroup$ 4 $\begingroup$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.
$\endgroup$ 1 $\begingroup$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 $x<0$
$\endgroup$ $\begingroup$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.
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