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Chapter 10: Problem 10

Find all of the exact solutions of the equation and then list those solutions which are in the interval \([0,2 \pi)\). $$ \cos \left(x+\frac{5 \pi}{6}\right)=0 $$

Short Answer

Expert verified
The solutions in the interval \([0, 2\pi)\) are \( x = \frac{2\pi}{3} \).

Step by step solution

01

Understand the Equation Form

We start with the equation \( \cos \left( x + \frac{5\pi}{6} \right) = 0 \). The cosine function equals zero at angles of the form \( \frac{(2k+1)\pi}{2} \), where \( k \) is an integer.
02

Solve for x + 5蟺/6

To solve for \( x + \frac{5\pi}{6} \), set \( x + \frac{5\pi}{6} = \frac{(2k+1)\pi}{2} \). This leads to a general solution: \( x = \frac{(2k+1)\pi}{2} - \frac{5\pi}{6} \).
03

Simplify the General Solution

Simplify \( x = \frac{(2k+1)\pi}{2} - \frac{5\pi}{6} \). Find a common denominator (6 in this case), resulting in \( x = \frac{3(2k+1)\pi}{6} - \frac{5\pi}{6} = \frac{(6k+3)\pi}{6} - \frac{5\pi}{6} = \frac{(6k-2)\pi}{6} \).
04

Solve for Values of k

To find \( x \) in the interval \([0, 2\pi)\), calculate \( \frac{(6k-2)\pi}{6} \) for different integer \( k \) values. Set \( 0 \leq \frac{(6k-2)\pi}{6} < 2\pi \), leading to \( 0 \leq 6k - 2 < 12 \). Solving gives integer \( k \), yielding specific \( x \) values.
05

Calculate and List x Solutions

Determine specific solutions, checking \( k = 0, 1, 2 \), etc. For \( k = 0 \), \( x = -\frac{2\pi}{6} = -\frac{\pi}{3} \), not in range. For \( k = 1 \), \( x = \frac{4\pi}{6} = \frac{2\pi}{3} \). For larger \( k \), results get out of range of \([0, 2\pi)\) or repeat.

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Key Concepts

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

Cosine Function
The cosine function is a fundamental trigonometric function that is commonly used to describe phenomena involving periodicity and waves. It represents the x-coordinate of a point on the unit circle, corresponding to a given angle.
The equation \( \cos \theta = 0 \) occurs when the angle \( \theta \) corresponds to the x-values where the circle crosses the y-axis.
  • These occur at angles \( \pi/2 \) and \( 3\pi/2 \).
  • For the cosine to be zero, the input (angle) must land on these points: \( \theta = \frac{(2k+1)\pi}{2} \), where \( k \) is an integer.
This periodic nature of the cosine function is why we often express solutions in terms of \( k \), allowing us to account for all potential solutions given by their rotations on the unit circle.
General Solution
A general solution in trigonometry refers to the expression of all possible angles that satisfy a given trigonometric equation. It captures the infinite set of angles that can meet the criteria set by the equation.
  • For the equation \( \cos \left(x + \frac{5\pi}{6}\right) = 0 \), we derive a general solution to account for all the angles at which the cosine function equals zero.
  • This specific arrangement leads us to find \( x + \frac{5\pi}{6} = \frac{(2k+1)\pi}{2} \).
  • Reworking the equation to solve for \( x \), we get \( x = \frac{(6k-2)\pi}{6} \). This expression accounts for all operative angles for every integer \( k \).
These encompass the solutions where any multiple of the period of the cosine (which is \( 2\pi \)) is added to the initial angle where cosine normally vanishes.
Angular Intervals
Working with angular intervals means identifying specific angle measurements that satisfy the constraints of a trigonometric problem within a defined range. The interval \([0, 2\pi)\) is particularly common and represents the range from 0 radians to just below full rotation.
  • The problem specifies that we must find the solutions within this interval, requiring us to consider angle values that complete zero to just under a complete circle.
  • Given the general solution \( x = \frac{(6k-2)\pi}{6} \), we determine values for \( k \) that yield \( x \) within the interval.
  • Solving the inequality \( 0 \leq 6k - 2 < 12 \) derives integer values for \( k \) as 1, which we use to find exact solutions.
  • This leads to valid \( x \) such as \( \frac{2\pi}{3} \), which lies squarely within the required range.
Exploring angular intervals is essential when we want to restrict our solutions to realistic and meaningful values within a specific cycle of periodicity.

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Most popular questions from this chapter

In Exercises 222 - 233 , find the domain of the given function. Write your answers in interval notation. $$ f(x)=\operatorname{arcsec}\left(\frac{x^{3}}{8}\right) $$

In Exercises \(131-154\), find the exact value or state that it is undefined. $$ \sin \left(\arccos \left(-\frac{1}{2}\right)\right) $$

In Exercises 222 - 233 , find the domain of the given function. Write your answers in interval notation. $$ f(x)=\operatorname{arccsc}(x+5) $$

In Exercises \(131-154\), find the exact value or state that it is undefined. $$ \cot (\operatorname{arccsc}(\sqrt{5})) $$

In Exercises \(131-154\), find the exact value or state that it is undefined. $$ \sec \left(\arcsin \left(-\frac{12}{13}\right)\right) $$
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