What is the number of possible $x$ values in $\frac{x}{100}= \sin(x)$ [duplicate]

Problem $\frac{x}{100}= \sin(x)$ We are asked to find the number of possible values of $x$ in this scenario and I had tried to figure it out by the use of trigonometric identities but then i had realized that there are no trig identities that can help me... or is there?

steps that i had tried: I first multiplied 100 on both sides to get $x$ by itself to get: $100(\sin x)=x$

Then i determined that $\sin x \le 1$ therefore I had determined that $\frac{x}{100}$ is $\le 1$

But when i look at the answer choices all of them are less than $100$ meaning that all of them are going to make $\frac{x}{100}$ less than 100 when $x$ is substituted.

marked as duplicate by user99914, user228113, Em., Community♦Jul 12 '16 at 6:08

• @angelo mark I understand that you are a perfectionist but maybe changing all the x's into $x$ is a bit too much? – John Rawls Jul 11 '16 at 7:25
• Sorry. I'll remember your username for not editing in future. Really Sorry. – Angelo Mark Jul 11 '16 at 7:55
• This question has already been answered. See here. – Bernard Jul 11 '16 at 8:12
• @AngeloMark don't worry man, i do the same and win 2 points. Lol – julio godoy Jul 11 '16 at 9:20

Since $|\sin x|\le 1$, one needs only to consider $\frac{|x|}{100} \le 1$, or $|x|\le 100$. Since both functions are odd, we restrict ourselves to the interval $[0,100]$. The curve

$$\tag{1} y = \frac{ x}{100}$$

is positive on $(0,100]$. $\sin x$ is periodic with period $2\pi$. On each (half) period

$$[0,\pi], [2\pi, 3\pi], \cdots [2n\pi, (2n+1)\pi]$$

as long as $(2n+1) \pi <100$, the curve $(1)$ intersects $\sin x$ at $2$ points. Since $15.5< 100/2\pi <16$, this means that there are $32$ nonnegative solution and thus $63$ solution in the real line.

• Are you did not make any errors? because the solution choices are 61, 62, 63. 64, or 65 – John Rawls Jul 11 '16 at 7:45
• @JohnRawls Yes, I suck at counting. – user99914 Jul 11 '16 at 7:48
• i'm sure you are right but can u explain on how u set 15<100/2π<16 and why this means that there are 32 nonnegative solutions and 63 real solutions? – John Rawls Jul 11 '16 at 7:51
• @JohnRawls : $15<100/2\pi<16$ are obtained by direct calculation (indeed I need $15.5 <100/2\pi$. This implies that there are $16$ such $[2n\pi, (2n+1)\pi]$ in $[0,100]$ and each gives me $2$ solutions. Thus we have $32$ in $[0,100]$. There are also $32$ solutions in $[-100,0]$, but that $0$ is in both counting, so the total number should be $32\times 2 -1 = 63$. – user99914 Jul 11 '16 at 7:55
• Really, I just plug in $100/2\pi$ in wolfram alpha and got around 15.9. – user99914 Jul 11 '16 at 7:58

look at this site body must be [1]: https://www.wolframalpha.com/input/?i=x%2F100%20%3D%20sin(x)