olve the equation.

sine 3 x equals StartFraction StartRoot 3 EndRoot Over 2 EndFractionsin3x=32​, 

0less than or equals≤xless than<2piπ

Question content area bottom

Part 1

Choose the correct answer below.

A.

xequals=StartFraction pi Over 12 EndFraction comma StartFraction 7 pi Over 12 EndFraction comma StartFraction 3 pi Over 4 EndFraction comma StartFraction 11 pi Over 12 EndFraction comma StartFraction 13 pi Over 12 EndFraction comma StartFraction 17 pi Over 12 EndFractionπ12,7π12,3π4,11π12,13π12,17π12

B.

xequals=StartFraction pi Over 12 EndFraction comma StartFraction pi Over 4 EndFraction comma StartFraction 3 pi Over 4 EndFraction comma StartFraction 11 pi Over 12 EndFraction comma StartFraction 17 pi Over 12 EndFraction comma StartFraction 19 pi Over 12 EndFractionπ12,π4,3π4,11π12,17π12,19π12

C.

xequals=StartFraction pi Over 9 EndFraction comma StartFraction 2 pi Over 9 EndFraction comma StartFraction 7 pi Over 9 EndFraction comma StartFraction 8 pi Over 9 EndFraction comma StartFraction 13 pi Over 9 EndFraction comma StartFraction 14 pi Over 9 EndFractionπ9,2π9,7π9,8π9,13π9,14π9

D.

xequals=StartFraction pi Over 9 EndFraction comma StartFraction 4 pi Over 9 EndFraction comma StartFraction 8 pi Over 9 EndFraction comma StartFraction 11 pi Over 9 EndFraction comma StartFraction 15 pi Over 9 EndFraction comma StartFraction 17 pi Over 9 EndFraction

Let's solve the equation \(\sin(3x) = \frac{\sqrt{3}}{2}\) in the interval \(0 \leq x < 2\pi\).

### Step 1: Determine the angles for \(3x\)

The equation \(\sin(3x) = \frac{\sqrt{3}}{2}\) corresponds to the standard angles where the sine function has this value. These angles are:

\[
3x = \frac{\pi}{3}, \frac{2\pi}{3}
\]

### Step 2: General solutions for \(3x\)

Since sine is periodic with a period of \(2\pi\), the general solutions for \(3x\) are:

\[
3x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad 3x = \frac{2\pi}{3} + 2k\pi \quad \text{for integers } k
\]

This can be simplified to:

\[
x = \frac{\pi}{9} + \frac{2k\pi}{3} \quad \text{and} \quad x = \frac{2\pi}{9} + \frac{2k\pi}{3}
\]

### Step 3: Generate specific solutions for \(x\)

To find all solutions for \(x\) within the interval \(0 \leq x < 2\pi\), let's substitute values of \(k\) to find valid \(x\):

- For \(3x = \frac{\pi}{3}\):

  - \(k = 0\): \(x = \frac{\pi}{9}\)
  - \(k = 1\): \(x = \frac{7\pi}{9}\)
  - \(k = 2\): \(x = \frac{13\pi}{9}\)
  
- For \(3x = \frac{2\pi}{3}\):

  - \(k = 0\): \(x = \frac{2\pi}{9}\)
  - \(k = 1\): \(x = \frac{8\pi}{9}\)
  - \(k = 2\): \(x = \frac{14\pi}{9}\)

### Step 4: List the solutions

The solutions \(x\) in the interval \(0 \leq x < 2\pi\) are:

\[
x = \frac{\pi}{9}, \frac{2\pi}{9}, \frac{7\pi}{9}, \frac{8\pi}{9}, \frac{13\pi}{9}, \frac{14\pi}{9}
\]

### Final Answer:

The correct answer is **Option C**:

\[
x = \frac{\pi}{9}, \frac{2\pi}{9}, \frac{7\pi}{9}, \frac{8\pi}{9}, \frac{13\pi}{9}, \frac{14\pi}{9}
\]

Would you like more details or have any other questions?

Here are 5 related questions:
1. How do you determine the general solutions for trigonometric equations?
2. What are the periods of sine and cosine functions, and how do they affect solutions?
3. How can you determine all solutions of a trigonometric equation within a given interval?
4. What is the relationship between the unit circle and solving trigonometric equations?
5. How can transformations of trigonometric functions affect the solutions?

**Tip:** When solving trigonometric equations like \(\sin(3x)\), remember to account for the periodicity of the function and check all possible values within the specified interval.

Learn how to solve the trigonometric equation sin(3x) = sqrt(3)/2 within the interval 0 <= x < 2π. Understand the properties of the sine function and its periodic nature. Suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation