MethodMath
WT
May 10, 2026

How to find the number of solutions to sin x = x/10 graphically?

This is an IIT-JEE problem that has been bothering me:

Find the number of solutions to the equation sinx=x10\sin x = \frac{x}{10} for xRx \in \mathbb{R}.

I tried solving it analytically but it seems impossible due to the transcendental nature. My teacher suggested a graphical approach. Could someone explain how to count the intersections methodically?

I know that both functions are odd, so the number of positive solutions will be symmetric to negative ones, with x=0x = 0 always being a solution.

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1 Answer

PD
Prof. David KimMay 10, 2026 Accepted
This is a classic problem that tests graphical visualization. Here is the systematic approach. **Step 1: Note symmetry.** Both $\sin x$ and $x/10$ are odd functions, so solutions come in symmetric pairs $\pm x$ plus $x = 0$. **Step 2: Bound the search region.** Since $|\sin x| \leq 1$, any solution must satisfy $|x/10| \leq 1$, i.e., $|x| \leq 10$. **Step 3: Count intersections in $[0, 10]$.** The line $y = x/10$ has slope $0.1$. The sine wave completes $\lfloor 10/(2\pi) \ floor = 1$ full period in $[0, 2\pi]$, then reaches halfway through the second period by $x = 10$. In each half-period where $\sin x$ is positive (i.e., $(0, \pi)$, $(2\pi, 3\pi)$), the line intersects the sine curve twice — once on the rising edge and once on the falling edge — provided the line's value at the peak exceeds the sine peak. Check the peaks: - At $x = \pi/2 \approx 1.57$, $\sin(\pi/2) = 1 > (\pi/2)/10 \approx 0.157$ ✓ - At $x = 2\pi + \pi/2 \approx 7.85$, $\sin = 1 > 0.785$ ✓ In the negative half-periods $(\pi, 2\pi)$ and $(3\pi, 10)$, the sine is negative, and the line is positive or crossing zero, yielding one intersection each. **Step 4: Count them.** Positive side: $x = 0$ counts once. We get 3 positive intersections (one in $(0,\pi)$, one in $(\pi,2\pi)$, one in $(2\pi,3\pi)$). By symmetry, there are 3 negative intersections. **Total: $1 + 3 + 3 = 7$ solutions.** This problem illustrates why transcendental equations often require graphical reasoning.
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