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Chapter 4: Problem 24

Find all real numbers that satisfy each equation. $$ 2 \sin (x)+\sqrt{3}=0 $$

Short Answer

Expert verified
The solutions are \( \frac{4\pi}{3} + 2k\pi \) and \( \frac{5\pi}{3} + 2k\pi \), where \( k \) is any integer.

Step by step solution

01

Isolate the Trigonometric Function

Subtract \( \sqrt{3} \) from both sides to isolate \( \sin(x) \): \ \ 2 \sin(x) = -\sqrt{3} \.
02

Solve for \( \sin(x) \)

Divide both sides by 2 to solve for \( \sin(x) \): \ \ \sin(x) = -\frac{\sqrt{3}}{2} \.
03

Determine the General Solution

Recall that \( \sin(x) = -\frac{\sqrt{3}}{2} \) at \( \frac{4\pi}{3} + 2k\pi \) and \( \frac{5\pi}{3} + 2k\pi \), 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.

Isolating Trigonometric Functions
To solve trigonometric equations, the first crucial step is to isolate the trigonometric function. This often involves rearranging the equation to get the trigonometric term by itself on one side. Let's see how to do it with our example:

We start with the equation:
\[ 2 \sin (x) + \sqrt{3} = 0 \]
Subtract \sqrt{3} from both sides to isolate the term with the sine function:
\[ 2 \sin(x) = -\sqrt{3} \]
Next, divide both sides by 2 to solve for \sin(x):
\[ \sin(x) = -\frac{\sqrt{3}}{2} \]
By getting \sin(x) alone, we make it easier to determine where and how frequently the trigonometric function equals this value. This is fundamental in solving these types of equations.
General Solution for Sine
The general solution for a trig function describes all possible solutions. Sine and cosine functions are periodic, meaning they repeat their values in regular intervals. For the sine function, it's crucial to remember where specific values occur within these periods.

For instance, \sin(x) = -\frac{\sqrt{3}}{2} happens:
\[ x = \frac{4\pi}{3} + 2k\pi \] and
\[ x = \frac{5\pi}{3} + 2k\pi \]
here, k is any integer representing how far along we are in the cycles.

This accounts for the periodicity of the sine function and means there are an infinite number of solutions that satisfy these equations. By adding multiples of the period (2\pi), we describe all possible solutions.
Solving Trigonometric Equations
Solving trigonometric equations combines these steps: isolating the trig function and using its general solution. Let's follow through with our example step by step:

  • Step 1: Isolate \sin(x):
    Subtract \sqrt{3} to get \ sin(x) by itself:
    \[ 2\sin(x) = -\sqrt{3} \]
  • Step 2: Solve for \sin(x):
    Divide both sides by 2:
    \[ \sin(x) = -\frac{\sqrt{3}}{2} \]
  • Step 3: Find where \sin(x) equals this value in its cycle:
    \[ x = \frac{4\pi}{3} + 2k\pi \]

    \[ x = \frac{5\pi}{3} + 2k\pi \]

Combining these steps ensures we address the nature of the trigonometric function and find all its solutions properly. Especially important is understanding the periodicity of sine and cosine for general solutions.

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Most popular questions from this chapter

A sector of a circle has a central angle of \(\pi / 6 .\) Find the exact area of the sector if the radius of the circle is 6 inches.

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