---
title: "Bisection: 3x - √(1+sin x)"
subject: "Numerical Method"
chapter: 2
key: "semester_4/numerical_method:bisection-3x-sqrt"
source: https://www.bctengineer.com/q/semester_4/numerical_method/bisection-3x-sqrt
site: BCT Engineering
---

# Bisection: 3x - √(1+sin x)

Numerical Method · Chapter 2: Solutions of Nonlinear Equations

- Asked in: 2072 Asoj, 8 marks
- Video: [|| f(x) = 3x - √(1+sin x) [0,1] up to three decimal places (thirteen iteration || class 12 ||](https://www.youtube.com/watch?v=uQPEInnLC4I) — @mathwithabbas
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/bisection-3x-sqrt
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-2

## Question

Using the Bisection method, find a real root of the equation f(x) = 3x - √(1+sin x) correct up to three decimal points.

## Short answer

Bisection with a=0,b=1 (f(0)<0,f(1)>0). After 11 iterations, interval width <0.001, giving root 0.392.

## 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) = 3x − √(1 + sin x)  (radians)
f(0) = −1 < 0 and f(1) = 3 − √1.8415 = 1.643 > 0, so a root lies in (0, 1).
Take a = 0, b = 1.

#### Step 2: Iterations
| n | a | b | c=(a+b)/2 | f(c) | root lies in |
|---|---|---|---|---|---|
| 1 | 0.0000 | 1.0000 | 0.5000 | 0.2837 | [a, c] |
| 2 | 0.0000 | 0.5000 | 0.2500 | -0.3669 | [c, b] |
| 3 | 0.2500 | 0.5000 | 0.3750 | -0.0439 | [c, b] |
| 4 | 0.3750 | 0.5000 | 0.4375 | 0.1193 | [a, c] |
| 5 | 0.3750 | 0.4375 | 0.4062 | 0.0376 | [a, c] |
| 6 | 0.3750 | 0.4062 | 0.3906 | -0.0032 | [c, b] |
| 7 | 0.3906 | 0.4062 | 0.3984 | 0.0172 | [a, c] |
| 8 | 0.3906 | 0.3984 | 0.3945 | 0.0070 | [a, c] |
| 9 | 0.3906 | 0.3945 | 0.3926 | 0.0019 | [a, c] |
| 10 | 0.3906 | 0.3926 | 0.3916 | -0.0006 | [c, b] |
| 11 | 0.3916 | 0.3926 | 0.3921 | 0.0006 | [a, c] |

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

**Answer: root ≈ 0.392**

## Check yourself

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

1. Root in (0, 1) because:
   A. f(0) = -1 < 0 and f(1) = 1.643 > 0
   B. f(0) > 0
   C. f(1) < 0
   D. f(0.5) = 0
   Answer: A. There is a sign change. (Key point: Bracket)

2. First midpoint is 0.5 with f = 0.2837 > 0, so the root lies in:
   A. [0.5, 1]
   B. [0, 0.5]
   C. [1, 2]
   D. At 0.5
   Answer: B. f(a)*f(c) < 0, so b = c. (Key point: First step)

3. Iterations stop after 11 steps since:
   A. f(c) = 0 exactly
   B. Width reached 0.1
   C. Interval width 0.00049 < 0.001
   D. A fixed rule says 11
   Answer: C. Three decimals need width below 0.001. (Key point: Stopping)

4. The root is:
   A. 0.500
   B. 0.250
   C. 0.400
   D. 0.392
   Answer: D. Midpoints converge near 0.392. (Key point: Result)

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