tan²(0) – 5 tan(0) + 6 = 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 Statement:**
Algebraically find all solutions to the following equation in the interval \([0^\circ, 360^\circ)\), rounding all solutions to the nearest \(0.01^\circ\).
\[
\tan^2(\theta) - 5\tan(\theta) + 6 = 0
\]
**Explanation:**
To solve this equation, we first consider it as a quadratic equation in terms of \(\tan(\theta)\). This quadratic can be expressed as:
\[
u^2 - 5u + 6 = 0
\]
where \(u = \tan(\theta)\).
**Solution Steps:**
1. **Factor the Quadratic:**
The quadratic equation can be factored into:
\[
(u - 3)(u - 2) = 0
\]
2. **Solve for \(u\):**
Set each factor to zero:
\[
u - 3 = 0 \quad \text{or} \quad u - 2 = 0
\]
This gives:
\[
u = 3 \quad \text{and} \quad u = 2
\]
3. **Find \(\theta\):**
Since \(u = \tan(\theta)\), find \(\theta\) for each value of \(u\).
- For \(u = 3\): \(\theta = \tan^{-1}(3)\)
- For \(u = 2\): \(\theta = \tan^{-1}(2)\)
4. **Consider all solutions in \([0^\circ, 360^\circ)\):**
Use the periodicity of tangent (\(180^\circ\) period) to find all solutions.
5. **Round the Solutions:**
After calculating \(\theta\) for both values of \(u\) and adding \(180^\circ n\) where \(n\) is an integer, round the results to the nearest \(0.01^\circ\) within the interval \([0^\circ, 360^\circ)\).
With these steps, you can find all solutions for \(\theta\) that satisfy the initial trigonometric equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1680afc1-b9ed-4705-8271-5c5463bd69a2%2Fc585bb28-25f2-4c85-aab9-4ad5601ae37f%2Fj58cdn_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Algebraically find all solutions to the following equation in the interval \([0^\circ, 360^\circ)\), rounding all solutions to the nearest \(0.01^\circ\).
\[
\tan^2(\theta) - 5\tan(\theta) + 6 = 0
\]
**Explanation:**
To solve this equation, we first consider it as a quadratic equation in terms of \(\tan(\theta)\). This quadratic can be expressed as:
\[
u^2 - 5u + 6 = 0
\]
where \(u = \tan(\theta)\).
**Solution Steps:**
1. **Factor the Quadratic:**
The quadratic equation can be factored into:
\[
(u - 3)(u - 2) = 0
\]
2. **Solve for \(u\):**
Set each factor to zero:
\[
u - 3 = 0 \quad \text{or} \quad u - 2 = 0
\]
This gives:
\[
u = 3 \quad \text{and} \quad u = 2
\]
3. **Find \(\theta\):**
Since \(u = \tan(\theta)\), find \(\theta\) for each value of \(u\).
- For \(u = 3\): \(\theta = \tan^{-1}(3)\)
- For \(u = 2\): \(\theta = \tan^{-1}(2)\)
4. **Consider all solutions in \([0^\circ, 360^\circ)\):**
Use the periodicity of tangent (\(180^\circ\) period) to find all solutions.
5. **Round the Solutions:**
After calculating \(\theta\) for both values of \(u\) and adding \(180^\circ n\) where \(n\) is an integer, round the results to the nearest \(0.01^\circ\) within the interval \([0^\circ, 360^\circ)\).
With these steps, you can find all solutions for \(\theta\) that satisfy the initial trigonometric equation.
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