---
title: "Bisection: sin x = 1/x"
subject: "Numerical Method"
chapter: 2
key: "semester_4/numerical_method:bisection-sinx-1-over-x"
source: https://www.bctengineer.com/q/semester_4/numerical_method/bisection-sinx-1-over-x
site: BCT Engineering
---

# Bisection: sin x = 1/x

Numerical Method · Chapter 2: Solutions of Nonlinear Equations

- Asked in: 2071 Bhadra, 8 marks
- Video: [Bisection Method Example | Numerical Methods](https://www.youtube.com/watch?v=Mk-HAyI3aoc) — StudySession
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/bisection-sinx-1-over-x
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-2

## Question

Use bisection method to find a real positive root of sin x = 1/x correct up to three decimal places.

## Short answer

With f(x)=sin x -1/x, bracket [1,2]. Biseection yields root 1.114 after 11 iterations (interval width <0.001).

## Answer

**Bisection method:** if f(a)·f(b) < 0, the root lies in [a, b]. Take c = (a + b)/2; if f(a)·f(c) < 0 the root is in [a, c], otherwise in [c, b]. Repeat until the interval is shorter than the required accuracy.

#### Step 1: Bracket the root
f(x) = sin x − 1/x  (radians)
f(1) = 0.8415 − 1 = −0.1585 < 0 and f(2) = 0.9093 − 0.5 = 0.4093 > 0, so a root lies in (1, 2).
Take a = 1, b = 2.

#### Step 2: Iterations
| n | a | b | c=(a+b)/2 | f(c) | root lies in |
|---|---|---|---|---|---|
| 1 | 1.0000 | 2.0000 | 1.5000 | 0.3308 | [a, c] |
| 2 | 1.0000 | 1.5000 | 1.2500 | 0.1490 | [a, c] |
| 3 | 1.0000 | 1.2500 | 1.1250 | 0.0134 | [a, c] |
| 4 | 1.0000 | 1.1250 | 1.0625 | -0.0676 | [c, b] |
| 5 | 1.0625 | 1.1250 | 1.0938 | -0.0259 | [c, b] |
| 6 | 1.0938 | 1.1250 | 1.1094 | -0.0060 | [c, b] |
| 7 | 1.1094 | 1.1250 | 1.1172 | 0.0038 | [a, c] |
| 8 | 1.1094 | 1.1172 | 1.1133 | -0.0011 | [c, b] |
| 9 | 1.1133 | 1.1172 | 1.1152 | 0.0013 | [a, c] |
| 10 | 1.1133 | 1.1152 | 1.1143 | 0.0001 | [a, c] |
| 11 | 1.1133 | 1.1143 | 1.1138 | -0.0005 | [c, b] |

After 11 iterations the interval width is 0.00049 (< 0.001), so the root is fixed to 3 decimals.

**Answer: root ≈ 1.114**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. sin x = 1/x is written as:
   A. f(x) = sin x - 1/x
   B. sin x + 1/x
   C. x sin x
   D. sin x - x
   Answer: A. Move all terms to one side. (Key point: Function)

2. The interval (1, 2) is used because:
   A. f(1) > 0
   B. f(1) = -0.1585 < 0, f(2) = 0.4093 > 0
   C. f(2) < 0
   D. f(1.5) = 0
   Answer: B. Sign change. (Key point: Bracket)

3. At c = 1.5, f = 0.3308 > 0; next interval is:
   A. [1.5, 2]
   B. [1.5, 1.5]
   C. [1, 1.5]
   D. [0, 1]
   Answer: C. f(a) < 0, f(c) > 0, so b = c. (Key point: Update)

4. The root is:
   A. 1.500
   B. 1.125
   C. 0.886
   D. 1.114
   Answer: D. Midpoints converge to 1.114. (Key point: Result)

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