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Sagot :
To solve the equation [tex]\( 2 \cos \left(\frac{x}{2}\right) + 1 = 0 \)[/tex] within the interval [tex]\( 0 \leq x < 2\pi \)[/tex], we need to follow several steps carefully.
Starting with the given equation:
[tex]\[ 2 \cos \left( \frac{x}{2} \right) + 1 = 0 \][/tex]
First, isolate the cosine function:
[tex]\[ \cos \left( \frac{x}{2} \right) = -\frac{1}{2} \][/tex]
Next, we need to determine the values of [tex]\( \frac{x}{2} \)[/tex] for which [tex]\( \cos \left( \frac{x}{2} \right) = -\frac{1}{2} \)[/tex]. Recall from the unit circle that:
[tex]\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \quad \text{and} \quad \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \][/tex]
Thus:
[tex]\[ \frac{x}{2} = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad \frac{x}{2} = \frac{4\pi}{3} + 2k\pi \][/tex]
for any integer [tex]\(k\)[/tex].
Multiplying both sides by 2 to solve for [tex]\(x\)[/tex], we get:
[tex]\[ x = \frac{4\pi}{3} + 4k\pi \quad \text{and} \quad x = \frac{8\pi}{3} + 4k\pi \][/tex]
We need to find solutions within the interval [tex]\( [0, 2\pi) \)[/tex]. Thus we check the candidates [tex]\(k = -1, 0, 1\)[/tex] to find valid [tex]\(x\)[/tex]:
For [tex]\(k = 0\)[/tex]:
- [tex]\( x = \frac{4\pi}{3} \)[/tex]
- [tex]\( x = \frac{8\pi}{3} \)[/tex] (which exceeds [tex]\(2\pi\)[/tex])
For [tex]\(k = -1\)[/tex]:
- [tex]\( x = \frac{4\pi}{3} - 4\pi = -\frac{8\pi}{3} \)[/tex] (invalid since it's negative)
- [tex]\( x = \frac{8\pi}{3} - 4\pi = \frac{5\pi}{3} \)[/tex]
For [tex]\(k = 1\)[/tex]:
- [tex]\( x = \frac{4\pi}{3} + 4\pi = \frac{16\pi}{3} \)[/tex] (which exceeds [tex]\(2\pi\)[/tex])
- [tex]\( x = \frac{8\pi}{3} + 4\pi = \frac{20\pi}{3} \)[/tex] (which exceeds [tex]\(2\pi\)[/tex])
Within the interval [tex]\( [0, 2\pi) \)[/tex], the valid solutions we have found are:
[tex]\[ x = \frac{4\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3} \][/tex]
Thus, the correct set of solutions for [tex]\(2 \cos \left(\frac{x}{2} \right) + 1 = 0\)[/tex] in the interval [tex]\( 0 \leq x < 2\pi \)[/tex] is:
[tex]\[ \left\{ \frac{4\pi}{3}, \frac{5\pi}{3} \right\} \][/tex]
So, the correct choices from the given options are:
[tex]\[ \left\{\frac{4 \pi}{3}\right\} \quad \text{and} \quad \left(\frac{5 \pi}{3}\right) \][/tex]
Starting with the given equation:
[tex]\[ 2 \cos \left( \frac{x}{2} \right) + 1 = 0 \][/tex]
First, isolate the cosine function:
[tex]\[ \cos \left( \frac{x}{2} \right) = -\frac{1}{2} \][/tex]
Next, we need to determine the values of [tex]\( \frac{x}{2} \)[/tex] for which [tex]\( \cos \left( \frac{x}{2} \right) = -\frac{1}{2} \)[/tex]. Recall from the unit circle that:
[tex]\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \quad \text{and} \quad \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \][/tex]
Thus:
[tex]\[ \frac{x}{2} = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad \frac{x}{2} = \frac{4\pi}{3} + 2k\pi \][/tex]
for any integer [tex]\(k\)[/tex].
Multiplying both sides by 2 to solve for [tex]\(x\)[/tex], we get:
[tex]\[ x = \frac{4\pi}{3} + 4k\pi \quad \text{and} \quad x = \frac{8\pi}{3} + 4k\pi \][/tex]
We need to find solutions within the interval [tex]\( [0, 2\pi) \)[/tex]. Thus we check the candidates [tex]\(k = -1, 0, 1\)[/tex] to find valid [tex]\(x\)[/tex]:
For [tex]\(k = 0\)[/tex]:
- [tex]\( x = \frac{4\pi}{3} \)[/tex]
- [tex]\( x = \frac{8\pi}{3} \)[/tex] (which exceeds [tex]\(2\pi\)[/tex])
For [tex]\(k = -1\)[/tex]:
- [tex]\( x = \frac{4\pi}{3} - 4\pi = -\frac{8\pi}{3} \)[/tex] (invalid since it's negative)
- [tex]\( x = \frac{8\pi}{3} - 4\pi = \frac{5\pi}{3} \)[/tex]
For [tex]\(k = 1\)[/tex]:
- [tex]\( x = \frac{4\pi}{3} + 4\pi = \frac{16\pi}{3} \)[/tex] (which exceeds [tex]\(2\pi\)[/tex])
- [tex]\( x = \frac{8\pi}{3} + 4\pi = \frac{20\pi}{3} \)[/tex] (which exceeds [tex]\(2\pi\)[/tex])
Within the interval [tex]\( [0, 2\pi) \)[/tex], the valid solutions we have found are:
[tex]\[ x = \frac{4\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3} \][/tex]
Thus, the correct set of solutions for [tex]\(2 \cos \left(\frac{x}{2} \right) + 1 = 0\)[/tex] in the interval [tex]\( 0 \leq x < 2\pi \)[/tex] is:
[tex]\[ \left\{ \frac{4\pi}{3}, \frac{5\pi}{3} \right\} \][/tex]
So, the correct choices from the given options are:
[tex]\[ \left\{\frac{4 \pi}{3}\right\} \quad \text{and} \quad \left(\frac{5 \pi}{3}\right) \][/tex]
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