/*! 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 47 \(2 \sin ^{2} \theta-3 \sin \the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 47

\(2 \sin ^{2} \theta-3 \sin \theta=-1\)

Short Answer

Expert verified
The solutions are \(\theta = \frac{\pi}{2}, \frac{\pi}{6}, 5\frac{\pi}{6}\) plus any multiple of \(2\pi\).

Step by step solution

01

Transform the Trigonometric Equation

First, let's rewrite the given equation in terms of a standard quadratic form. Start with \(2 \sin^2 \theta - 3 \sin \theta = -1\), and add 1 to both sides to get: \(2 \sin^2 \theta - 3 \sin \theta + 1 = 0\).
02

Substitution for Simplification

Let \(x = \sin \theta\). This changes the equation to: \(2x^2 - 3x + 1 = 0\).
03

Solve the Quadratic Equation

We will solve the quadratic equation \(2x^2 - 3x + 1 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 2\), \(b = -3\), and \(c = 1\).
04

Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\). Substituting the values, \((-3)^2 - 4(2)(1) = 9 - 8 = 1\). Since the discriminant is positive, there are two real solutions.
05

Use the Quadratic Formula

Substitute the calculated values into the quadratic formula: \(x = \frac{-(-3) \pm \sqrt{1}}{2 \times 2}\), which simplifies to \(x = \frac{3 \pm 1}{4}\).
06

Find the Values of x

Calculate the two potential solutions for \(x\):- \(x = \frac{3 + 1}{4} = 1\)- \(x = \frac{3 - 1}{4} = \frac{1}{2}\)
07

Determine the Trigonometric Solutions

Now revert back to \(\sin \theta = x\) with possible values \(\sin \theta = 1\) or \(\frac{1}{2}\). - For \(\sin \theta = 1\), \(\theta = \frac{\pi}{2} + 2n\pi\), where \(n\) is an integer,- For \(\sin \theta = \frac{1}{2}\), \(\theta = \frac{\pi}{6} + 2n\pi\) or \(5\frac{\pi}{6} + 2n\pi\), 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!

Key Concepts

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

Understanding Trigonometric Identities
Trigonometric identities are essential tools in simplifying and solving equations that involve trigonometric functions like sine, cosine, and tangent. They allow us to transform expressions into more manageable forms.
One of the fundamental identities is the Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\).
  • This identity relates sine and cosine, serving as a basis for many transformations.
  • By rearranging it, you can express \(\sin^2 \theta\) as \(1 - \cos^2 \theta\) or \(\cos^2 \theta\) as \(1 - \sin^2 \theta\).
These identities are also useful in combining and converting between trigonometric functions. Applying them wisely can simplify an equation like \(2 \sin^2 \theta - 3 \sin \theta = -1\) into a standard quadratic form for easier solving.
Exploring Sinusoidal Functions
Sinusoidal functions, represented by sine and cosine waves, play a crucial role in modeling periodic phenomena. These functions are characterized by their amplitude, period, and phase.
  • Amplitude: This is the peak value of the wave from its rest position. For \(\sin \theta\), the standard amplitude is 1, unless modified by a coefficient.
  • Period: Defined as the distance over which the function repeats itself. For sine and cosine, the standard period is \(2\pi\).
  • Phase Shift: This determines where the wave starts. Adding or subtracting a constant to the angle inside the function shifts it horizontally.
By understanding these properties, you can determine specific values and behavior of the waveforms at any given point, critical for solving trigonometric equations using solutions like \(\sin \theta = \frac{1}{2}\).
Solving Quadratic Equations Using the Quadratic Formula
Quadratic equations can often appear in trigonometry, especially when solving for angles using functions like sine and cosine. The quadratic formula is a powerful method for finding solutions to these equations:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
  • Coefficients: In the general quadratic equation \(ax^2 + bx + c = 0\), \(a\), \(b\), and \(c\) are the coefficients.
  • Discriminant: The expression \(b^2 - 4ac\) is known as the discriminant. The value of the discriminant determines the number and nature of the solutions:
    • If it's positive, there are two distinct real solutions.
    • If it's zero, there is one real solution.
    • If it's negative, there are no real solutions, but two complex ones.
Using the quadratic formula, you can find the roots of equations like \(2x^2 - 3x + 1 = 0\), which simplifies the problem of finding angle values such as \(\theta\) achieved in trigonometric equations.

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

The opposite of a vector is a vector with the ______ magnitude and __________ direction. To obtain the opposite of a vector, multiply the vector by ____.

Force A chair lift at a ski resort is stopped halfway between two poles that support the cable to which the chair is attached. The poles are 215 feet apart and the combined weight of the chair and the three people on the chair is 725 pounds. If the weight of the chair and the people riding it causes the chair to move to a position \(15.8\) feet below the horizontal line that connects the top of the two poles, find the tension in the cable toward each end of the cable.

$$ \text { Solve each of the following triangles. } $$ $$ \text { Use the law of cosines to show that, if } A=90^{\circ} \text {, then } a^{2}=b^{2}+c^{2} \text {. } $$

Distance A hot-air balloon is held at a constant altitude by two ropes that are anchored to the ground. One rope is 120 feet long and makes an angle of \(65^{\circ}\) with the ground. The other rope is 115 feet long. What is the distance between the points on the ground at which the two ropes are anchored?

For Problems 37 through 42, use your knowledge of bearing, heading, and true course to sketch a diagram that will help you solve each problem. Heading and Distance Two planes take off at the same time from an airport. The first plane is flying at 246 miles per hour on a course of \(135.0^{\circ}\). The second plane is flying in the direction \(175.0^{\circ}\) at 357 miles per hour. Assuming there are no wind currents blowing, how far apart are they after 2 hours?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Calculus

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.