Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
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]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
