The equation $x-\sin x = 0$ If we have the equation $x-\sin x=0$, then we can trivially or numerically find the solution to be $x=0$. However, I rearrange the equation algebraically and get $$\frac{\sin x}x=1.$$ 
If I plug $x=0$ into $\displaystyle \frac{\sin x}x$, I don't get a $1$; it's actually undefined.
(And yes, I know, $\displaystyle \lim_{x\rightarrow 0} \frac{\sin x}x=1$. Not sure if this helps though.)
 A: You should write since you divide by $x$ which it vanishes at $0$:
$$\sin x-x=0\iff \left(\frac{\sin x}x=1\right)\lor \left(x=0\right)$$
A: Your equation should be:
$$x = \sin x$$
Only solution occurs is $x = 0$.
A: First define $f(x) = x - \sin(x)$ and note that $f(0) = 0 - \sin(0) = 0$. Now, consider $x>0$. Then $f'(x) = 1 - \cos(x) \geq 0$ and so $f(x) \geq 0$. Also note that $$x - \sin(x) \geq \frac{x^3}6 - \frac{x^5}{120} \geq \frac{x^3}6 - \frac{x^3}{120} = \frac{19x^3}{120} > 0,$$ for all $x \in (0,1].$ Then, since $f$ is non-decreasing, $f(x) \geq f(1) > 0$, $\forall x \in (1, \infty)$. The other case for $x<0$ is similar.
A: Since you tagged "precalculus" i'll do my best to give a simple answer. The first equation is not "equivalent" to the second equation, in the sense that they dont have the same set of solutions. This appears strange to you, because you got to the second one by doing simple algebraic manipulations to the first. And you ask the question: "What gives?". 
The answer is: If you believe that one equation is equivalent to the other the you should prove it (note that this is how real math is done, not like in school where you are simply taught to manipulate variables). In this case, as you found out, this is not possible since in the first eqn 0 is obviously a solution, and the other is not defined at that point. So the real lesson here is (not what you expected i guess) that manipulating equations doesn't necessarily get you equivalent equations. In this case you devided by a quantity that you then set equal to zero. Obviously this is trouble...
The last answer given by Sami Ben Romdhane is also correct but very short and i thought i would elaborate a bit more.
Hope this helps.
