Find all solutions of the equation in the interval [0, 27). cos 4x cos2x+ sin 4x sin2x = 0 Write your answer in radians in terms of . If there is more than one solution, separate them with commas. x =
Find all solutions of the equation in the interval [0, 27). cos 4x cos2x+ sin 4x sin2x = 0 Write your answer in radians in terms of . If there is more than one solution, separate them with commas. x =
Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter2: Right Triangle Trigonometry
Section: Chapter Questions
Problem 6GP
Related questions
Question
Find all solutions of the equation
![### Solving Trigonometric Equations: Cosine and Sine Functions
#### Problem Statement
Find all solutions of the equation in the interval \([0, 2\pi)\):
\[ \cos 4x \cos 2x + \sin 4x \sin 2x = 0 \]
#### Instructions
Write your answer in radians in terms of \(\pi\).
If there is more than one solution, separate them with commas.
\[ x = \]
#### Explanation of Equations
The equation given combines the cosine and sine functions in a trigonometric identity. To solve this, we can use the sum-to-product identities which simplify trigonometric expressions. Specifically, for the given equation, we can apply the identity:
\[ \cos A \cos B + \sin A \sin B = \cos(A - B) \]
Given:
\[ \cos 4x \cos 2x + \sin 4x \sin 2x = 0 \]
This matches the form of the identity with \(A = 4x\) and \(B = 2x\):
\[ \cos(4x - 2x) = \cos 2x \]
Thus, the equation simplifies to:
\[ \cos 2x = 0 \]
Now, let's find the values of \(x\) in the interval \([0, 2\pi)\) where \(\cos 2x = 0\).
1. \(\cos 2x = 0\) implies \(2x = \frac{\pi}{2} + k\pi\).
2. Solving for \(x\), we get:
\[ x = \frac{\pi}{4} + \frac{k\pi}{2} \]
where \(k\) is an integer.
#### Finding Solutions
The principal values of \(x\) need to be found within the interval \([0, 2\pi)\):
1. For \(k = 0\):
\[ x = \frac{\pi}{4} \]
2. For \(k = 1\):
\[ x = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \]
3. For \(k = 2\):
\[ x = \frac{\pi}{4} + \pi = \frac{5\pi}{4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c0c699c-0ea3-4621-a196-a76a2f333ed8%2Ff4177ab6-1db4-4fbd-b119-5505bf2017e1%2F4l027j_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving Trigonometric Equations: Cosine and Sine Functions
#### Problem Statement
Find all solutions of the equation in the interval \([0, 2\pi)\):
\[ \cos 4x \cos 2x + \sin 4x \sin 2x = 0 \]
#### Instructions
Write your answer in radians in terms of \(\pi\).
If there is more than one solution, separate them with commas.
\[ x = \]
#### Explanation of Equations
The equation given combines the cosine and sine functions in a trigonometric identity. To solve this, we can use the sum-to-product identities which simplify trigonometric expressions. Specifically, for the given equation, we can apply the identity:
\[ \cos A \cos B + \sin A \sin B = \cos(A - B) \]
Given:
\[ \cos 4x \cos 2x + \sin 4x \sin 2x = 0 \]
This matches the form of the identity with \(A = 4x\) and \(B = 2x\):
\[ \cos(4x - 2x) = \cos 2x \]
Thus, the equation simplifies to:
\[ \cos 2x = 0 \]
Now, let's find the values of \(x\) in the interval \([0, 2\pi)\) where \(\cos 2x = 0\).
1. \(\cos 2x = 0\) implies \(2x = \frac{\pi}{2} + k\pi\).
2. Solving for \(x\), we get:
\[ x = \frac{\pi}{4} + \frac{k\pi}{2} \]
where \(k\) is an integer.
#### Finding Solutions
The principal values of \(x\) need to be found within the interval \([0, 2\pi)\):
1. For \(k = 0\):
\[ x = \frac{\pi}{4} \]
2. For \(k = 1\):
\[ x = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \]
3. For \(k = 2\):
\[ x = \frac{\pi}{4} + \pi = \frac{5\pi}{4
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