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Chapter 10: Problem 9

Solve each equation for solutions over the interval \([0,2 \pi)\) by first solving for the trigonometric finction. Do not use a calculator. $$2 \csc x+4=\csc x+6$$

Short Answer

Expert verified
The solutions are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).

Step by step solution

01

Isolate the Trigonometric Function

First, let's start by isolating the trigonometric function in the equation. We have: \[2 \csc x + 4 = \csc x + 6\]Subtract \( \csc x \) from both sides to get: \[csc x + 4 = 6\]
02

Solve for the Cosecant

Next, isolate \( \csc x \) by subtracting 4 from both sides: \[\csc x = 2\]
03

Convert Cosecant to Sine

Remember that \( \csc x = \frac{1}{\sin x} \). Knowing this, set up the equation: \[\frac{1}{\sin x} = 2\]Solve for \( \sin x \) by taking the reciprocal: \[\sin x = \frac{1}{2}\]
04

Find Solutions in the Given Interval

We need to find \( x \) such that \( \sin x = \frac{1}{2} \) within the interval \([0, 2\pi)\). The sine function equals \( \frac{1}{2} \) at \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \). Therefore, the solutions are: \[x = \frac{\pi}{6}, \frac{5\pi}{6}\]

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Key Concepts

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

Cosecant Function
The cosecant function, denoted as \( \csc x \), is one of the reciprocal trigonometric functions. It's closely related to the sine function, since \( \csc x \) is defined as \( \frac{1}{\sin x} \). This means that wherever the sine function is defined and not equal to zero, the cosecant function will also be defined.
  • When \( \sin x = 1 \), \( \csc x = 1 \).
  • When \( \sin x = \frac{1}{2} \), \( \csc x = 2 \).
It's important to note that when \( \sin x = 0 \), the value of \( \csc x \) becomes undefined, because division by zero is not possible. Thus, the cosecant is crucial for problems where solutions involve angles corresponding to specific values of sine.
Sine Function
The sine function, represented as \( \sin x \), is one of the fundamental trigonometric functions. It indicates the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. The sine function is periodic, repeating its values every \( 2\pi \).
  • The values of \( \sin x \) range from -1 to 1.
  • It reaches 1 at \( x = \frac{\pi}{2} + 2k\pi \), where \( k \) is an integer.
  • It reaches -1 at \( x = \frac{3\pi}{2} + 2k\pi \).
  • It equals 0 at \( x = k\pi \).
In solving trigonometric equations, converting expressions into terms of sine can simplify finding solutions within a given interval. Like in the exercise, finding when \( \sin x = \frac{1}{2} \) helps pinpoint the angle values that fulfill the trigonometric equation.
Interval Notation
Interval notation is a concise way of expressing a range of numbers, often used in mathematics to specify the domain where a function or equation is applicable. In the context of trigonometric equations, it tells us where we should search for valid solutions. In this exercise, the interval is given as \([0, 2\pi)\).
  • The \([\) indicates that \( 0 \) is included in the interval.
  • The \(()\) indicates that \( 2\pi \) is not included.
Using interval notation helps to systematically determine which angles fall within the allowable range for a trigonometric solution. It鈥檚 especially useful when dealing with periodic functions like sine, as it helps limit solutions to those within the specified cycle.

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Most popular questions from this chapter

Use a calculator to find each value. $$\cot (\arccos 0.58236841)$$

Verify that each equation is an identity by using any of the identities introduced in the first three sections of this chapter. $$\frac{1-\sin t}{\cos t}=\frac{1}{\sec t+\tan t}$$

Solve each problem. Hearing Beats in Music Musicians sometimes tune instruments by playing the same tone on two different instruments and listening for a phenomenon known as beats. When the two instruments are in tune, the beats disappear. The ear hears beats because the pressure slowly rises and falls as a result of the slight variation in the frequency. This phenomenon can be seen on a graphing calculator. (a) Consider two tones with frequencies of 220 and \(223 \mathrm{Hz}\) and pressures \(P_{1}=0.005 \sin 440 \pi t \quad\) and \(\quad P_{2}=0.005 \sin 446 \pi t\) respectively. A graph of \(P_{1}+P_{2}\) as \(Y_{3}\) felt by an eardrum over the 1 -second interval \([0.15,1.15]\) is shown here. How many beats are there in 1 second? (Graph can't copy) (b) Repeat part (a) with frequencies of 220 and 216 (c) Determine a simple way to find the number of beats per second if the frequency of each tone is given.

Solve each equation ( \(x\) in radians and \(\theta\) in degrees) for all exact solutions where appropriate. Round approximale values in radians to four decimal places and approximate values in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$\sin 2 x=2 \cos ^{2} x$$

Solve each equation ( \(x\) in radians and \(\theta\) in degrees) for all exact solutions where appropriate. Round approximale values in radians to four decimal places and approximate values in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$2 \cos ^{2} 2 \theta=1-\cos 2 \theta$$
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