olve the equation 8cos8+4=10. Express the answer in degree measure and approximate the angular measure to the nearest enth of a degree. mny 48.6°+360°k or 131.4° + 360° k 41.4°+360°k or 138.6° + 360°k 41.4°+360°k or 318.6° + 360° k 138.6°+360°k or 221.4° +360°k 20 Q Search † 4:17 PM 6/25/2023
olve the equation 8cos8+4=10. Express the answer in degree measure and approximate the angular measure to the nearest enth of a degree. mny 48.6°+360°k or 131.4° + 360° k 41.4°+360°k or 138.6° + 360°k 41.4°+360°k or 318.6° + 360° k 138.6°+360°k or 221.4° +360°k 20 Q Search † 4:17 PM 6/25/2023
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Solving the Equation 8cosθ + 4 = 10**
**Instructions:**
Solve the equation \(8cosθ + 4 = 10\). Express the answer in degree measure and approximate the angular measure to the nearest tenth of a degree.
**Multiple Choice Options:**
1. \(48.6^\circ + 360^\circ k \quad or \quad 131.4^\circ + 360^\circ k\)
2. \(41.4^\circ + 360^\circ k \quad or \quad 138.6^\circ + 360^\circ k\) ⬅️ This is selected.
3. \(41.4^\circ + 360^\circ k \quad or \quad 318.6^\circ + 360^\circ k\)
4. \(138.6^\circ + 360^\circ k \quad or \quad 221.4^\circ + 360^\circ k\)
**Explanation:**
Given the equation:
\[8cosθ + 4 = 10\]
First, isolate \(cosθ\):
\[8cosθ = 10 - 4\]
\[8cosθ = 6\]
\[cosθ = \frac{6}{8}\]
\[cosθ = \frac{3}{4}\]
Next, we find the angle θ whose cosine is \( \frac{3}{4} \). We use the arccos function:
\[θ = arccos\left(\frac{3}{4}\right)\]
This calculation gives us two primary solutions within one full rotation (0° to 360°):
\[θ_1 ≈ 41.4^\circ\]
\[θ_2 = 360° - 41.4° ≈ 318.6^\circ\]
Since the cosine function is periodic with a period of 360°, these solutions repeat every 360°:
\[θ_1 = 41.4^\circ + 360^\circ k\]
\[θ_2 = 318.6^\circ + 360^\circ k\]
Therefore, the correct selection should be:
\[41.4^\circ + 360^\circ k \quad or \quad 318.6^\circ + 360^\circ k\]
So the closest match in the given options is:
**Option 3.**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f02248d-9b04-4bf9-ba6b-18c74681389d%2F4784f43c-8bc7-408e-b1fc-9dc184146359%2Feokcrmf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Solving the Equation 8cosθ + 4 = 10**
**Instructions:**
Solve the equation \(8cosθ + 4 = 10\). Express the answer in degree measure and approximate the angular measure to the nearest tenth of a degree.
**Multiple Choice Options:**
1. \(48.6^\circ + 360^\circ k \quad or \quad 131.4^\circ + 360^\circ k\)
2. \(41.4^\circ + 360^\circ k \quad or \quad 138.6^\circ + 360^\circ k\) ⬅️ This is selected.
3. \(41.4^\circ + 360^\circ k \quad or \quad 318.6^\circ + 360^\circ k\)
4. \(138.6^\circ + 360^\circ k \quad or \quad 221.4^\circ + 360^\circ k\)
**Explanation:**
Given the equation:
\[8cosθ + 4 = 10\]
First, isolate \(cosθ\):
\[8cosθ = 10 - 4\]
\[8cosθ = 6\]
\[cosθ = \frac{6}{8}\]
\[cosθ = \frac{3}{4}\]
Next, we find the angle θ whose cosine is \( \frac{3}{4} \). We use the arccos function:
\[θ = arccos\left(\frac{3}{4}\right)\]
This calculation gives us two primary solutions within one full rotation (0° to 360°):
\[θ_1 ≈ 41.4^\circ\]
\[θ_2 = 360° - 41.4° ≈ 318.6^\circ\]
Since the cosine function is periodic with a period of 360°, these solutions repeat every 360°:
\[θ_1 = 41.4^\circ + 360^\circ k\]
\[θ_2 = 318.6^\circ + 360^\circ k\]
Therefore, the correct selection should be:
\[41.4^\circ + 360^\circ k \quad or \quad 318.6^\circ + 360^\circ k\]
So the closest match in the given options is:
**Option 3.**
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