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Chapter 8: Problem 25

Find all solutions of the equation. $$2 \sin 3 x+1=0$$

Short Answer

Expert verified
The solutions are \(x = -\frac{\pi}{18} + \frac{2k\pi}{3}\) and \(x = \frac{7\pi}{18} + \frac{2k\pi}{3}\), where \(k\) is any integer.

Step by step solution

01

Isolate the Sine Function

The given equation is \(2 \sin 3x + 1 = 0\). First, isolate the sine function by subtracting 1 from both sides: \(2 \sin 3x = -1\). Then, divide by 2: \(\sin 3x = -\frac{1}{2}\).
02

Find General Solution for Sine

The equation \(\sin \theta = -\frac{1}{2}\) has solutions \(\theta = -\frac{\pi}{6} + 2k\pi\) and \(\theta = \frac{7\pi}{6} + 2k\pi\), where \(k\) is any integer. Therefore, the general solutions for \(3x\) are: \(3x = -\frac{\pi}{6} + 2k\pi\) and \(3x = \frac{7\pi}{6} + 2k\pi\).
03

Solve for x

Divide each of the solutions for \(3x\) by 3 to find \(x\): From \(3x = -\frac{\pi}{6} + 2k\pi\), we get \(x = -\frac{\pi}{18} + \frac{2k\pi}{3}\).From \(3x = \frac{7\pi}{6} + 2k\pi\), we get \(x = \frac{7\pi}{18} + \frac{2k\pi}{3}\).
04

Write the Final General Solution

The solutions for \(x\) are: \(x = -\frac{\pi}{18} + \frac{2k\pi}{3}\) and \(x = \frac{7\pi}{18} + \frac{2k\pi}{3}\), where \(k\) is any integer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

The Sine Function
The sine function is a fundamental trigonometric function that arises in the context of right-angle triangles and waveforms. It represents the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. However, in deeper contexts like unit circles, it is defined as the y-coordinate of a point on the unit circle corresponding to a given angle. This means:
  • It oscillates between -1 and 1.
  • It has a period of \(2\pi\), meaning the shape repeats every \(2\pi\) radians.
In the given equation, we start with \(2 \sin 3x + 1 = 0\). First, we isolate the sine term to form \(\sin 3x = -\frac{1}{2}\). This indicates that \(3x\) is an angle where the sine has a specific value, part of the inverse sine process.
General Solution of Sine Equations
Solving a trigonometric equation like the one given involves finding the "general solution." This solution represents all possible angles or values of \(x\) that satisfy the equation. For sine functions, the general approach includes finding the initial solution and then adding integral multiples of the period to capture all possible solutions. Specifically, if \(\sin \theta = a\), the general solutions can be expressed as \(\theta = \arcsin(a) + 2k\pi\) or \(\theta = \pi - \arcsin(a) + 2k\pi\).
In this problem:
  • The sine of \(\theta = -\frac{1}{2}\) leads to the solutions \(\theta = -\frac{\pi}{6} + 2k\pi\) and \(\theta = \frac{7\pi}{6} + 2k\pi\) since these angles lie in the third and fourth quadrants where sine is negative.
Thus, for \(3x\), the general solutions are derived based on these outcomes.
Finding Angle Solutions for x
The process of finding angle solutions for \(x\) requires us to adjust the general solutions of the given equation to isolate \(x\). Starting from an equation like \(3x = \theta\):
  • We divide every part of the solution by 3 to account for the factor inside the sine function argument, transforming \(\theta\) into \(x\).
For example:
  • Start with \(3x = -\frac{\pi}{6} + 2k\pi\), divide by 3: \(x = -\frac{\pi}{18} + \frac{2k\pi}{3}\).
  • Another option \(3x = \frac{7\pi}{6} + 2k\pi\), divide by 3: \(x = \frac{7\pi}{18} + \frac{2k\pi}{3}\).
These angle solutions fully describe all possible values for \(x\) that satisfy the original trigonometric equation because \(k\) can be any integer, encompassing all the cycles described by the general solution.

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