Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To solve the equation [tex]\(\cos(2x) = \cos(x)\)[/tex] over the interval [tex]\([0, 2\pi)\)[/tex], we can follow these steps:
1. Use the double-angle identity for cosine:
Recall the double-angle identity for cosine:
[tex]\[ \cos(2x) = 2\cos^2(x) - 1 \][/tex]
Using this, the equation [tex]\(\cos(2x) = \cos(x)\)[/tex] becomes:
[tex]\[ 2\cos^2(x) - 1 = \cos(x) \][/tex]
2. Rearrange the equation to form a quadratic equation in terms of [tex]\(\cos(x)\)[/tex]:
[tex]\[ 2\cos^2(x) - \cos(x) - 1 = 0 \][/tex]
3. Solve the quadratic equation:
Let [tex]\(u = \cos(x)\)[/tex]. The equation becomes:
[tex]\[ 2u^2 - u - 1 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -1\)[/tex]:
[tex]\[ u = \frac{1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} \][/tex]
This gives us two solutions for [tex]\(u\)[/tex]:
[tex]\[ u = \frac{4}{4} = 1 \quad \text{and} \quad u = \frac{-2}{4} = -\frac{1}{2} \][/tex]
Therefore, [tex]\(\cos(x) = 1\)[/tex] or [tex]\(\cos(x) = -\frac{1}{2}\)[/tex].
4. Find the corresponding values of [tex]\(x\)[/tex]:
For [tex]\(\cos(x) = 1\)[/tex]:
[tex]\[ x = 0 \quad (\text{within the interval } [0, 2\pi)) \][/tex]
For [tex]\(\cos(x) = -\frac{1}{2}\)[/tex]:
[tex]\[ x = \frac{2\pi}{3} \quad \text{and} \quad x = \frac{4\pi}{3} \quad (\text{within the interval } [0, 2\pi)) \][/tex]
5. Check given potential solutions:
Given potential solutions to check are:
[tex]\[ 0, \quad \frac{\pi}{3}, \quad \frac{2\pi}{3}, \quad \pi, \quad \frac{4\pi}{3}, \quad \frac{5\pi}{3} \][/tex]
Comparing with our solutions [tex]\(\{0, \frac{2\pi}{3}, \frac{4\pi}{3}\}\)[/tex]:
- 0: Is a solution.
- [tex]\(\frac{\pi}{3}\)[/tex]: Is not a solution.
- [tex]\(\frac{2\pi}{3}\)[/tex]: Is a solution.
- [tex]\(\pi\)[/tex]: Is not a solution (as [tex]\(\cos(\pi) = -1\)[/tex], not [tex]\(-\frac{1}{2}\)[/tex] or 1).
- [tex]\(\frac{4\pi}{3}\)[/tex]: Is a solution.
- [tex]\(\frac{5\pi}{3}\)[/tex]: Is not a solution (as [tex]\(\cos(\frac{5\pi}{3}) = \frac{1}{2}\)[/tex], not [tex]\(-\frac{1}{2}\)[/tex] or 1).
Therefore, the solutions to [tex]\(\cos(2x) = \cos(x)\)[/tex] over the interval [tex]\([0, 2\pi)\)[/tex], checking all provided options, are:
- 0
Thus, the only correct option given is:
[tex]\[ 0 \][/tex]
1. Use the double-angle identity for cosine:
Recall the double-angle identity for cosine:
[tex]\[ \cos(2x) = 2\cos^2(x) - 1 \][/tex]
Using this, the equation [tex]\(\cos(2x) = \cos(x)\)[/tex] becomes:
[tex]\[ 2\cos^2(x) - 1 = \cos(x) \][/tex]
2. Rearrange the equation to form a quadratic equation in terms of [tex]\(\cos(x)\)[/tex]:
[tex]\[ 2\cos^2(x) - \cos(x) - 1 = 0 \][/tex]
3. Solve the quadratic equation:
Let [tex]\(u = \cos(x)\)[/tex]. The equation becomes:
[tex]\[ 2u^2 - u - 1 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -1\)[/tex]:
[tex]\[ u = \frac{1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} \][/tex]
This gives us two solutions for [tex]\(u\)[/tex]:
[tex]\[ u = \frac{4}{4} = 1 \quad \text{and} \quad u = \frac{-2}{4} = -\frac{1}{2} \][/tex]
Therefore, [tex]\(\cos(x) = 1\)[/tex] or [tex]\(\cos(x) = -\frac{1}{2}\)[/tex].
4. Find the corresponding values of [tex]\(x\)[/tex]:
For [tex]\(\cos(x) = 1\)[/tex]:
[tex]\[ x = 0 \quad (\text{within the interval } [0, 2\pi)) \][/tex]
For [tex]\(\cos(x) = -\frac{1}{2}\)[/tex]:
[tex]\[ x = \frac{2\pi}{3} \quad \text{and} \quad x = \frac{4\pi}{3} \quad (\text{within the interval } [0, 2\pi)) \][/tex]
5. Check given potential solutions:
Given potential solutions to check are:
[tex]\[ 0, \quad \frac{\pi}{3}, \quad \frac{2\pi}{3}, \quad \pi, \quad \frac{4\pi}{3}, \quad \frac{5\pi}{3} \][/tex]
Comparing with our solutions [tex]\(\{0, \frac{2\pi}{3}, \frac{4\pi}{3}\}\)[/tex]:
- 0: Is a solution.
- [tex]\(\frac{\pi}{3}\)[/tex]: Is not a solution.
- [tex]\(\frac{2\pi}{3}\)[/tex]: Is a solution.
- [tex]\(\pi\)[/tex]: Is not a solution (as [tex]\(\cos(\pi) = -1\)[/tex], not [tex]\(-\frac{1}{2}\)[/tex] or 1).
- [tex]\(\frac{4\pi}{3}\)[/tex]: Is a solution.
- [tex]\(\frac{5\pi}{3}\)[/tex]: Is not a solution (as [tex]\(\cos(\frac{5\pi}{3}) = \frac{1}{2}\)[/tex], not [tex]\(-\frac{1}{2}\)[/tex] or 1).
Therefore, the solutions to [tex]\(\cos(2x) = \cos(x)\)[/tex] over the interval [tex]\([0, 2\pi)\)[/tex], checking all provided options, are:
- 0
Thus, the only correct option given is:
[tex]\[ 0 \][/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
