16. Given 0 0 2π, solve 2sin 03sine + 1 = 0
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
Related questions
Question
![## Problem 16
### Given \(0 \leq \theta \leq 2\pi\), solve:
\[
2\sin^2{\theta} - 3\sin{\theta} + 1 = 0
\]
### Solution:
To solve this trigonometric equation, follow these steps:
1. **Recognize the Quadratic Form:**
The equation \(2\sin^2{\theta} - 3\sin{\theta} + 1 = 0\) can be treated as a quadratic equation in terms of \(\sin{\theta}\). Let \(x = \sin{\theta}\).
So the equation becomes:
\[
2x^2 - 3x + 1 = 0
\]
2. **Solve the Quadratic Equation:**
To solve for \(x\), use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -3\), and \(c = 1\).
\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2}
\]
\[
x = \frac{3 \pm \sqrt{9 - 8}}{4}
\]
\[
x = \frac{3 \pm 1}{4}
\]
This gives us two solutions:
\[
x = \frac{3 + 1}{4} = 1
\]
\[
x = \frac{3 - 1}{4} = \frac{1}{2}
\]
So, \( \sin{\theta} = 1 \) or \( \sin{\theta} = \frac{1}{2} \).
3. **Determine the Angles \(\theta\):**
From \( \sin{\theta} = 1 \):
\[
\theta = \frac{\pi}{2}
\]
From \( \sin{\theta} = \frac{1}{2} \):
\[
\theta = \frac{\pi}{6}
\]
and
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F050b7896-96e1-4a53-ae28-da4cd34956eb%2Fbafa5aa1-fad9-4ec5-a747-0a65db61577d%2F52sgab_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Problem 16
### Given \(0 \leq \theta \leq 2\pi\), solve:
\[
2\sin^2{\theta} - 3\sin{\theta} + 1 = 0
\]
### Solution:
To solve this trigonometric equation, follow these steps:
1. **Recognize the Quadratic Form:**
The equation \(2\sin^2{\theta} - 3\sin{\theta} + 1 = 0\) can be treated as a quadratic equation in terms of \(\sin{\theta}\). Let \(x = \sin{\theta}\).
So the equation becomes:
\[
2x^2 - 3x + 1 = 0
\]
2. **Solve the Quadratic Equation:**
To solve for \(x\), use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -3\), and \(c = 1\).
\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2}
\]
\[
x = \frac{3 \pm \sqrt{9 - 8}}{4}
\]
\[
x = \frac{3 \pm 1}{4}
\]
This gives us two solutions:
\[
x = \frac{3 + 1}{4} = 1
\]
\[
x = \frac{3 - 1}{4} = \frac{1}{2}
\]
So, \( \sin{\theta} = 1 \) or \( \sin{\theta} = \frac{1}{2} \).
3. **Determine the Angles \(\theta\):**
From \( \sin{\theta} = 1 \):
\[
\theta = \frac{\pi}{2}
\]
From \( \sin{\theta} = \frac{1}{2} \):
\[
\theta = \frac{\pi}{6}
\]
and
\[
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