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Sagot :
To solve the equation [tex]\(\tan \theta - \sqrt{3} = 0\)[/tex] within the interval [tex]\([0, 2\pi)\)[/tex], follow these steps:
1. Rewrite the Equation:
[tex]\[ \tan \theta = \sqrt{3} \][/tex]
2. Find the Principal Solution:
We need to determine [tex]\(\theta\)[/tex] such that the tangent of [tex]\(\theta\)[/tex] equals [tex]\(\sqrt{3}\)[/tex]. Recall that:
[tex]\[ \tan \left(\frac{\pi}{3}\right) = \sqrt{3} \][/tex]
So, [tex]\(\theta = \frac{\pi}{3}\)[/tex] is one solution.
3. Consider the Periodicity of the Tangent Function:
The tangent function has a period of [tex]\(\pi\)[/tex]. This means that if [tex]\(\theta\)[/tex] is a solution, then [tex]\(\theta + k\pi\)[/tex] is also a solution for any integer [tex]\(k\)[/tex]. Hence, the general solution can be written as:
[tex]\[ \theta = \frac{\pi}{3} + k\pi \quad \text{for integer } k \][/tex]
4. Find Solutions within the Given Interval [tex]\([0, 2\pi)\)[/tex]:
We need to consider values of [tex]\(k\)[/tex] that keep [tex]\(\theta\)[/tex] within the interval [tex]\([0, 2\pi)\)[/tex].
- For [tex]\(k = 0\)[/tex]:
[tex]\[ \theta = \frac{\pi}{3} \][/tex]
Since [tex]\(\frac{\pi}{3}\)[/tex] is in [tex]\([0, 2\pi)\)[/tex], it is a valid solution.
- For [tex]\(k = 1\)[/tex]:
[tex]\[ \theta = \frac{\pi}{3} + \pi = \frac{\pi}{3} + \frac{3\pi}{3} = \frac{4\pi}{3} \][/tex]
Since [tex]\(\frac{4\pi}{3}\)[/tex] is in [tex]\([0, 2\pi)\)[/tex], it is also a valid solution.
- For [tex]\(k = 2\)[/tex]:
[tex]\[ \theta = \frac{\pi}{3} + 2\pi = \frac{\pi}{3} + \frac{6\pi}{3} = \frac{7\pi}{3} \][/tex]
Since [tex]\(\frac{7\pi}{3} > 2\pi\)[/tex], it is outside the interval [tex]\([0, 2\pi)\)[/tex] and hence not a valid solution.
Therefore, the solutions within the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[ \theta = \frac{\pi}{3} \quad \text{and} \quad \theta = \frac{4\pi}{3} \][/tex]
5. Converting Solutions to Decimal Form (if needed, for better understanding):
- [tex]\(\theta = \frac{\pi}{3} \approx 1.0471975511965976\)[/tex]
- [tex]\(\theta = \frac{4\pi}{3} \approx 4.1887902047863905\)[/tex]
Thus, the solutions to the equation [tex]\(\tan \theta - \sqrt{3} = 0\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[ \boxed{1.0471975511965976 \quad \text{and} \quad 4.1887902047863905} \][/tex]
1. Rewrite the Equation:
[tex]\[ \tan \theta = \sqrt{3} \][/tex]
2. Find the Principal Solution:
We need to determine [tex]\(\theta\)[/tex] such that the tangent of [tex]\(\theta\)[/tex] equals [tex]\(\sqrt{3}\)[/tex]. Recall that:
[tex]\[ \tan \left(\frac{\pi}{3}\right) = \sqrt{3} \][/tex]
So, [tex]\(\theta = \frac{\pi}{3}\)[/tex] is one solution.
3. Consider the Periodicity of the Tangent Function:
The tangent function has a period of [tex]\(\pi\)[/tex]. This means that if [tex]\(\theta\)[/tex] is a solution, then [tex]\(\theta + k\pi\)[/tex] is also a solution for any integer [tex]\(k\)[/tex]. Hence, the general solution can be written as:
[tex]\[ \theta = \frac{\pi}{3} + k\pi \quad \text{for integer } k \][/tex]
4. Find Solutions within the Given Interval [tex]\([0, 2\pi)\)[/tex]:
We need to consider values of [tex]\(k\)[/tex] that keep [tex]\(\theta\)[/tex] within the interval [tex]\([0, 2\pi)\)[/tex].
- For [tex]\(k = 0\)[/tex]:
[tex]\[ \theta = \frac{\pi}{3} \][/tex]
Since [tex]\(\frac{\pi}{3}\)[/tex] is in [tex]\([0, 2\pi)\)[/tex], it is a valid solution.
- For [tex]\(k = 1\)[/tex]:
[tex]\[ \theta = \frac{\pi}{3} + \pi = \frac{\pi}{3} + \frac{3\pi}{3} = \frac{4\pi}{3} \][/tex]
Since [tex]\(\frac{4\pi}{3}\)[/tex] is in [tex]\([0, 2\pi)\)[/tex], it is also a valid solution.
- For [tex]\(k = 2\)[/tex]:
[tex]\[ \theta = \frac{\pi}{3} + 2\pi = \frac{\pi}{3} + \frac{6\pi}{3} = \frac{7\pi}{3} \][/tex]
Since [tex]\(\frac{7\pi}{3} > 2\pi\)[/tex], it is outside the interval [tex]\([0, 2\pi)\)[/tex] and hence not a valid solution.
Therefore, the solutions within the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[ \theta = \frac{\pi}{3} \quad \text{and} \quad \theta = \frac{4\pi}{3} \][/tex]
5. Converting Solutions to Decimal Form (if needed, for better understanding):
- [tex]\(\theta = \frac{\pi}{3} \approx 1.0471975511965976\)[/tex]
- [tex]\(\theta = \frac{4\pi}{3} \approx 4.1887902047863905\)[/tex]
Thus, the solutions to the equation [tex]\(\tan \theta - \sqrt{3} = 0\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[ \boxed{1.0471975511965976 \quad \text{and} \quad 4.1887902047863905} \][/tex]
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