8. Solve the following equation for all x: cos 2x + sin x = 0 (no decimals)
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°.
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![**Title: Solving Trigonometric Equations**
**Problem Statement:**
8. Solve the following equation for all \( x \):
\[ \cos 2x + \sin x = 0 \]
*(Note: Answers should be given in exact form, no decimals.)*
---
**Solution:**
To solve the equation \( \cos 2x + \sin x = 0 \), we can use trigonometric identities and algebraic manipulation.
### Step 1: Use a Trigonometric Identity
Recall the double-angle identity for cosine:
\[ \cos 2x = 1 - 2\sin^2 x \quad \text{or} \quad \cos 2x = 2\cos^2 x - 1 \]
For the purpose of simplification, we will use the identity:
\[ \cos 2x = 1 - 2\sin^2 x \]
### Step 2: Substitute the Identity
Substitute \( \cos 2x \) in the original equation:
\[ 1 - 2\sin^2 x + \sin x = 0 \]
### Step 3: Form a Quadratic Equation
Rearrange the terms to form a quadratic equation in terms of \( \sin x \):
\[ -2\sin^2 x + \sin x + 1 = 0 \]
Multiply through by -1 to simplify the coefficients:
\[ 2\sin^2 x - \sin x - 1 = 0 \]
### Step 4: Solve the Quadratic Equation
We can solve this quadratic equation using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -1 \), and \( c = -1 \):
\[ \sin x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} \]
Simplify the expression under the square root:
\[ \sin x = \frac{1 \pm \sqrt{1 + 8}}{4} \]
\[ \sin x = \frac{1 \pm \sqrt{9}}{4} \]
\[ \sin x = \frac{1 \pm 3}{4} \]
This gives us two solutions:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd88595c7-c499-4476-ae94-4573188e2c46%2Fb28dfa91-dcf3-4f0e-89ba-2522cc6da878%2Fqhcbyf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Solving Trigonometric Equations**
**Problem Statement:**
8. Solve the following equation for all \( x \):
\[ \cos 2x + \sin x = 0 \]
*(Note: Answers should be given in exact form, no decimals.)*
---
**Solution:**
To solve the equation \( \cos 2x + \sin x = 0 \), we can use trigonometric identities and algebraic manipulation.
### Step 1: Use a Trigonometric Identity
Recall the double-angle identity for cosine:
\[ \cos 2x = 1 - 2\sin^2 x \quad \text{or} \quad \cos 2x = 2\cos^2 x - 1 \]
For the purpose of simplification, we will use the identity:
\[ \cos 2x = 1 - 2\sin^2 x \]
### Step 2: Substitute the Identity
Substitute \( \cos 2x \) in the original equation:
\[ 1 - 2\sin^2 x + \sin x = 0 \]
### Step 3: Form a Quadratic Equation
Rearrange the terms to form a quadratic equation in terms of \( \sin x \):
\[ -2\sin^2 x + \sin x + 1 = 0 \]
Multiply through by -1 to simplify the coefficients:
\[ 2\sin^2 x - \sin x - 1 = 0 \]
### Step 4: Solve the Quadratic Equation
We can solve this quadratic equation using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -1 \), and \( c = -1 \):
\[ \sin x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} \]
Simplify the expression under the square root:
\[ \sin x = \frac{1 \pm \sqrt{1 + 8}}{4} \]
\[ \sin x = \frac{1 \pm \sqrt{9}}{4} \]
\[ \sin x = \frac{1 \pm 3}{4} \]
This gives us two solutions:
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