/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Find all solutions of the equati... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find all solutions of the equation. $$\csc \gamma=\sqrt{2}$$

Short Answer

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

Step by step solution

01

Convert Cosecant to Sine

Recognize that the cosecant function is the reciprocal of the sine function. Thus, we have \( ext{csc} \, \gamma = \frac{1}{\sin \gamma}\). So, rewrite the equation in terms of sine: \(\frac{1}{\sin \gamma} = \sqrt{2}\).
02

Solve for Sine

From \(\frac{1}{\sin \gamma} = \sqrt{2}\), solve for \(\sin \gamma\). Multiply both sides by \(\sin \gamma\) and then divide by \(\sqrt{2}\) to get \(\sin \gamma = \frac{1}{\sqrt{2}}\).
03

Simplify Sine Expression

Rationalize the denominator of \(\sin \gamma = \frac{1}{\sqrt{2}}\) to get \(\sin \gamma = \frac{\sqrt{2}}{2}\).
04

Determine General Solutions for Sine

Recall that \(\sin \gamma = \frac{\sqrt{2}}{2}\) at angles \(\gamma = \frac{\pi}{4} + 2k\pi\) and \(\gamma = \frac{3\pi}{4} + 2k\pi\) for any integer \(k\). These represent the principal angles and the periodic nature of the sine function.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Cosecant Function
The cosecant function is one of the lesser-known trigonometric functions, but it plays a significant role in solving trigonometric equations. In simple terms, cosecant is the reciprocal of the sine function. This means that if you obtain the sine of an angle, the cosecant is the inverse value of that sine. Mathematically, it is represented as:
  • ext{csc} heta = rac{1}{ ext{sin} heta}
Understanding the reciprocal relationship here is crucial, especially when you need to work with equations involving the cosecant function. This relationship allows you to convert cosecant equations into sine equations, which are easier to solve.
Converting a problem from cosecant to sine can make it more approachable since sine functions have well-known properties and solutions.
Sine Function
The sine function is foundational in trigonometry and common in various applications, from geometry to calculus. It's a periodic function, meaning it repeats its values in regular intervals, specifically every \(2\pi\). The sine of an angle \(\theta\) can be defined as:
  • \( ext{sin} \, \theta = \frac{ ext{opposite side}}{ ext{hypotenuse}}\)
when dealing with a right triangle.
In the unit circle context, sine is the \(y\)-coordinate of a point where a line through the origin makes an angle \(\theta\) with the positive \(x\)-axis. Understanding how sine behaves and how it is linked to different angles can help solve trigonometric equations by finding angles that satisfy given conditions.
Reciprocal Trigonometric Identities
Reciprocal trigonometric identities are a set of identities that express each of the six trigonometric functions in terms of reciprocal relationships. These identities are crucial when solving trigonometric equations, like when we converted cosecant to sine in the exercise. Here are the fundamental reciprocal identities:
  • \( ext{csc} heta = \frac{1}{ ext{sin} heta}\)
  • \( ext{sec} heta = \frac{1}{ ext{cos} heta}\)
  • \( ext{cot} heta = \frac{1}{ ext{tan} heta}\)
Using these identities can simplify trigonometric expressions and help convert them into more familiar forms. This is particularly useful when solving equations that may initially seem complex.
General Solutions for Trigonometric Equations
General solutions in trigonometric equations refer to finding all possible solutions for an equation, given the periodic nature of trigonometric functions. Since trigonometric functions like sine repeat every \(2\pi\), there are typically infinite solutions for a given trigonometric equation, except within specific bounds.
For example, in the problem "\( ext{sin} \, \gamma = \frac{\sqrt{2}}{2}\)", the principal solutions are \(\gamma = \frac{\pi}{4}\) and \(\gamma = \frac{3\pi}{4}\).
However, since sine is periodic with a period of \(2\pi\), the complete solution set includes:
  • \(\gamma = \frac{\pi}{4} + 2k\pi\)
  • \(\gamma = \frac{3\pi}{4} + 2k\pi\)
where \(k\) is any integer. This general format accounts for all repetitions of the angle on the unit circle that have the same sine value.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Approximate the solution to each inequality on the interval \([0,2 \pi]\). $$\tan x \leq \sin 2 x$$

Sketch the graph of the equation. $$y=\frac{1}{2} \sin ^{-1} x$$

Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. \(15 \cos ^{4} x-14 \cos ^{2} x+3=0\) \([0, \pi]\)

Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. $$\begin{aligned} &2 \tan ^{2} t+9 \tan t+3=0\\\ &\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{aligned}$$

Pressure on the eardrum If a tuning fork is struck and then held a certain distance from the eardrum, the pressure \(p_{1}(t)\) on the outside of the eardrum at time \(t\) may be represented by \(p_{1}(t)=A\) sin \(\omega t,\) where \(A\) and \(\omega\) are positive constants. If a second identical tuning fork is struck with a possibly different force and held a different distance from the eardrum (see the figure), its effect may be represented by the equation \(p_{2}(t)=B \sin (\omega t+\tau),\) where \(B\) is a positive constant and \(0 \leq \tau \leq 2 \pi .\) The total pressure \(p(t)\) on the eardrum is given by $$ p(t)=A \sin \omega t+B \sin (\omega t+\tau) $$ (a) Show that \(p(t)=a \cos \omega t+b \sin \omega t,\) where $$ a=B \sin \tau \quad \text { and } \quad b=A+B \cos \tau $$ (b) Show that the amplitude \(C\) of \(p\) is given by $$ C^{2}=A^{2}+B^{2}+2 A B \cos \tau $$ (IMAGE CAN'T COPY)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.