/*! 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 938 Find the solution set of \(5 \ta... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 29: Problem 938

Find the solution set of \(5 \tan ^{2} \alpha-2 \tan \alpha-1=0\).

Short Answer

Expert verified
The solution set of the given quadratic equation \(5\tan^2{\alpha} -2\tan{\alpha} -1 = 0\) is given by: \(伪_1 = \arctan{\frac{2 + 2\sqrt{6}}{10}} + n\pi\) and \(伪_2 = \arctan{\frac{2 - 2\sqrt{6}}{10}} + n\pi\), where n is an integer.

Step by step solution

01

Identify the quadratic equation in terms of tan(伪)

The given equation is a quadratic equation in terms of \(\tan{\alpha}\): \(5\tan^2{\alpha} -2\tan{\alpha} -1 = 0\).
02

Find the roots of the quadratic equation

To find the roots of the quadratic equation, we apply the quadratic formula: \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). Here, \(a = 5\), \(b = -2\), and \(c = -1\). First, let's calculate the discriminant (螖): \(螖 = b^2-4ac = (-2)^2 - 4(5)(-1) = 24\). Now, we can find the roots: \(\tan{\alpha_1} = \frac{-(-2) + \sqrt{24}}{2(5)} = \frac{2 + 2\sqrt{6}}{10}\) and \(\tan{\alpha_2} = \frac{-(-2) - \sqrt{24}}{2(5)} = \frac{2 - 2\sqrt{6}}{10}\).
03

Find the solution set for 伪 in the range [0, 2蟺)

Now that we know the roots, we need to find the general solution for 伪 in the range [0, 2蟺). For \(\tan{\alpha_1} = \frac{2 + 2\sqrt{6}}{10}\), We take the arctangent of both sides: \(伪_1 = \arctan{\frac{2 + 2\sqrt{6}}{10}}\). For \(\tan{\alpha_2} = \frac{2 - 2\sqrt{6}}{10}\), We take the arctangent of both sides: \(伪_2 = \arctan{\frac{2 - 2\sqrt{6}}{10}}\).
04

Find the general solution

Finally, we can write the general solution set for 伪 in terms of the roots we found earlier. Since tan(伪) is periodic with period 蟺, the general solution will be: \(伪_1 = \arctan{\frac{2 + 2\sqrt{6}}{10}} + n\pi\) and \(伪_2 = \arctan{\frac{2 - 2\sqrt{6}}{10}} + n\pi\), where n is an integer.

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 91影视!

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

Prove that the following equation is an identity: \((1-\sin x) /(\cos x)=(\cos x) /(1+\sin x)\).

Find the solution set of \(2 \cos ^{2} x-5 \cos x+2=0\).

Prove the identity \(\csc 2 \mathrm{x}=(\csc \mathrm{x}) /(2 \cos \mathrm{x})\)

Prove \(\sin \left(45^{\circ}+\mathrm{x}\right)+\sin \left(45^{\circ}-\mathrm{x}\right)=\sqrt{2} \cos \mathrm{x}\)

Solve \(2 \sin ^{2} \theta+3 \cos \theta-3=0\) for \(\theta\) if \(0 \leq \theta<360^{\circ}\).
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation
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.