Answer :
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]
