Answer :
To solve the equation [tex]\(2 \cos^2 x + \cos x - 1 = 0\)[/tex] for [tex]\(x\)[/tex] over the interval [tex]\([0, 2\pi)\)[/tex], let us follow these steps:
1. Substitute [tex]\(u = \cos x\)[/tex]:
The equation becomes a quadratic in [tex]\(u\)[/tex]:
[tex]\[ 2u^2 + u - 1 = 0 \][/tex]
2. Solve the 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],
we have:
[tex]\[ u = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \][/tex]
[tex]\[ u = \frac{-1 \pm \sqrt{1 + 8}}{4} \][/tex]
[tex]\[ u = \frac{-1 \pm \sqrt{9}}{4} \][/tex]
[tex]\[ u = \frac{-1 \pm 3}{4} \][/tex]
This gives us two solutions for [tex]\(u\)[/tex]:
[tex]\[ u_1 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2} \][/tex]
[tex]\[ u_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1 \][/tex]
3. Find [tex]\(\cos x = \frac{1}{2}\)[/tex] solutions:
We need to find [tex]\(x\)[/tex] such that [tex]\(\cos x = \frac{1}{2}\)[/tex].
In the interval [tex]\([0, 2\pi)\)[/tex], this happens at:
[tex]\[ x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3} \][/tex]
4. Find [tex]\(\cos x = -1\)[/tex] solutions:
We need to find [tex]\(x\)[/tex] such that [tex]\(\cos x = -1\)[/tex].
In the interval [tex]\([0, 2\pi)\)[/tex], this happens at:
[tex]\[ x = \pi \][/tex]
5. List all solutions:
The solutions in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[ x = \frac{\pi}{3}, \pi, \frac{5\pi}{3} \][/tex]
Comparing these solutions with the given options:
- A. [tex]\( \pi \)[/tex] and [tex]\(\frac{\pi}{3}\)[/tex] – Not correct, it’s missing [tex]\(\frac{5\pi}{3}\)[/tex].
- B. [tex]\(\pi, \frac{\pi}{3}\)[/tex], and [tex]\(\frac{5\pi}{3}\)[/tex] – Correct, this matches our solutions.
- C. 1 and [tex]\(\frac{2\pi}{3}\)[/tex] – Not correct, these values do not satisfy our cosine equation.
- D. 1, [tex]\(\frac{2\pi}{3}\)[/tex], and [tex]\(\frac{4\pi}{3}\)[/tex] – Not correct, these values do not satisfy the cosine solutions.
- E. 1, [tex]\(\frac{\pi}{\pi}\)[/tex], and [tex]\(\frac{5\pi}{\pi}\)[/tex] – Not correct and not properly formatted.
Therefore, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
1. Substitute [tex]\(u = \cos x\)[/tex]:
The equation becomes a quadratic in [tex]\(u\)[/tex]:
[tex]\[ 2u^2 + u - 1 = 0 \][/tex]
2. Solve the 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],
we have:
[tex]\[ u = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \][/tex]
[tex]\[ u = \frac{-1 \pm \sqrt{1 + 8}}{4} \][/tex]
[tex]\[ u = \frac{-1 \pm \sqrt{9}}{4} \][/tex]
[tex]\[ u = \frac{-1 \pm 3}{4} \][/tex]
This gives us two solutions for [tex]\(u\)[/tex]:
[tex]\[ u_1 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2} \][/tex]
[tex]\[ u_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1 \][/tex]
3. Find [tex]\(\cos x = \frac{1}{2}\)[/tex] solutions:
We need to find [tex]\(x\)[/tex] such that [tex]\(\cos x = \frac{1}{2}\)[/tex].
In the interval [tex]\([0, 2\pi)\)[/tex], this happens at:
[tex]\[ x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3} \][/tex]
4. Find [tex]\(\cos x = -1\)[/tex] solutions:
We need to find [tex]\(x\)[/tex] such that [tex]\(\cos x = -1\)[/tex].
In the interval [tex]\([0, 2\pi)\)[/tex], this happens at:
[tex]\[ x = \pi \][/tex]
5. List all solutions:
The solutions in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[ x = \frac{\pi}{3}, \pi, \frac{5\pi}{3} \][/tex]
Comparing these solutions with the given options:
- A. [tex]\( \pi \)[/tex] and [tex]\(\frac{\pi}{3}\)[/tex] – Not correct, it’s missing [tex]\(\frac{5\pi}{3}\)[/tex].
- B. [tex]\(\pi, \frac{\pi}{3}\)[/tex], and [tex]\(\frac{5\pi}{3}\)[/tex] – Correct, this matches our solutions.
- C. 1 and [tex]\(\frac{2\pi}{3}\)[/tex] – Not correct, these values do not satisfy our cosine equation.
- D. 1, [tex]\(\frac{2\pi}{3}\)[/tex], and [tex]\(\frac{4\pi}{3}\)[/tex] – Not correct, these values do not satisfy the cosine solutions.
- E. 1, [tex]\(\frac{\pi}{\pi}\)[/tex], and [tex]\(\frac{5\pi}{\pi}\)[/tex] – Not correct and not properly formatted.
Therefore, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
