Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To solve the equation [tex]\(\tan^2(x) + 2\sqrt{3}\tan(x) + 3 = 0\)[/tex] for [tex]\(x\)[/tex] in the range [tex]\(\pi \leq x \leq 2\pi\)[/tex], we will follow these steps:
1. Solve the quadratic equation for [tex]\(\tan(x)\)[/tex]:
[tex]\[ t = \tan(x) \][/tex]
This simplifies the given equation to:
[tex]\[ t^2 + 2\sqrt{3}t + 3 = 0 \][/tex]
Here, we can solve for [tex]\(t\)[/tex] using the quadratic formula, [tex]\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 2\sqrt{3}\)[/tex], and [tex]\(c = 3\)[/tex].
Substituting these values into the quadratic formula:
[tex]\[ t = \frac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \][/tex]
[tex]\[ t = \frac{-2\sqrt{3} \pm \sqrt{12 - 12}}{2} \][/tex]
[tex]\[ t = \frac{-2\sqrt{3} \pm \sqrt{0}}{2} \][/tex]
[tex]\[ t = \frac{-2\sqrt{3}}{2} \][/tex]
[tex]\[ t = -\sqrt{3} \][/tex]
So, the solution for [tex]\(\tan(x)\)[/tex] is:
[tex]\[ \tan(x) = -\sqrt{3} \][/tex]
2. Find the corresponding [tex]\(x\)[/tex] values in the range [tex]\(\pi \leq x \leq 2\pi\)[/tex]:
We need to find [tex]\(x\)[/tex] such that:
[tex]\[ \tan(x) = -\sqrt{3} \][/tex]
To find these values, we look for the arctangent (or inverse tangent) solutions and consider the periodicity of the tangent function. The basic solution for [tex]\(\tan(x) = -\sqrt{3}\)[/tex] within one period can be found using:
[tex]\[ x = \arctan(-\sqrt{3}) + k\pi \][/tex]
where [tex]\( k \)[/tex] is an integer. Specifically,
[tex]\[ \arctan(-\sqrt{3}) \approx -\frac{\pi}{3} \][/tex]
3. Adjust for the given range [tex]\(\pi \leq x \leq 2\pi\)[/tex]:
For the solutions to fall within [tex]\(\pi \leq x \leq 2\pi\)[/tex], we consider adding [tex]\( k\pi \)[/tex] to our arctan solution from above.
- For [tex]\(k = 1\)[/tex]:
[tex]\[ x = -\frac{\pi}{3} + \pi = \frac{2\pi}{3} \][/tex]
This value does not fall within the specified range.
- For [tex]\(k = 2\)[/tex]:
[tex]\[ x = -\frac{\pi}{3} + 2\pi = \frac{5\pi}{3} \][/tex]
This value does fall within the range [tex]\(\pi \leq x \leq 2\pi\)[/tex].
Thus, the suitable [tex]\(x\)[/tex] value falling in the specified range is:
[tex]\[ x = \frac{5\pi}{3} \approx 5.236 \][/tex]
So, the complete solution indicates that the value of [tex]\(x\)[/tex] in the interval [tex]\(\pi \leq x \leq 2\pi\)[/tex] that satisfies the equation [tex]\(\tan^2(x) + 2\sqrt{3}\tan(x) + 3 = 0\)[/tex] is [tex]\( x = \frac{5\pi}{3} \approx 5.236 \)[/tex].
1. Solve the quadratic equation for [tex]\(\tan(x)\)[/tex]:
[tex]\[ t = \tan(x) \][/tex]
This simplifies the given equation to:
[tex]\[ t^2 + 2\sqrt{3}t + 3 = 0 \][/tex]
Here, we can solve for [tex]\(t\)[/tex] using the quadratic formula, [tex]\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 2\sqrt{3}\)[/tex], and [tex]\(c = 3\)[/tex].
Substituting these values into the quadratic formula:
[tex]\[ t = \frac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \][/tex]
[tex]\[ t = \frac{-2\sqrt{3} \pm \sqrt{12 - 12}}{2} \][/tex]
[tex]\[ t = \frac{-2\sqrt{3} \pm \sqrt{0}}{2} \][/tex]
[tex]\[ t = \frac{-2\sqrt{3}}{2} \][/tex]
[tex]\[ t = -\sqrt{3} \][/tex]
So, the solution for [tex]\(\tan(x)\)[/tex] is:
[tex]\[ \tan(x) = -\sqrt{3} \][/tex]
2. Find the corresponding [tex]\(x\)[/tex] values in the range [tex]\(\pi \leq x \leq 2\pi\)[/tex]:
We need to find [tex]\(x\)[/tex] such that:
[tex]\[ \tan(x) = -\sqrt{3} \][/tex]
To find these values, we look for the arctangent (or inverse tangent) solutions and consider the periodicity of the tangent function. The basic solution for [tex]\(\tan(x) = -\sqrt{3}\)[/tex] within one period can be found using:
[tex]\[ x = \arctan(-\sqrt{3}) + k\pi \][/tex]
where [tex]\( k \)[/tex] is an integer. Specifically,
[tex]\[ \arctan(-\sqrt{3}) \approx -\frac{\pi}{3} \][/tex]
3. Adjust for the given range [tex]\(\pi \leq x \leq 2\pi\)[/tex]:
For the solutions to fall within [tex]\(\pi \leq x \leq 2\pi\)[/tex], we consider adding [tex]\( k\pi \)[/tex] to our arctan solution from above.
- For [tex]\(k = 1\)[/tex]:
[tex]\[ x = -\frac{\pi}{3} + \pi = \frac{2\pi}{3} \][/tex]
This value does not fall within the specified range.
- For [tex]\(k = 2\)[/tex]:
[tex]\[ x = -\frac{\pi}{3} + 2\pi = \frac{5\pi}{3} \][/tex]
This value does fall within the range [tex]\(\pi \leq x \leq 2\pi\)[/tex].
Thus, the suitable [tex]\(x\)[/tex] value falling in the specified range is:
[tex]\[ x = \frac{5\pi}{3} \approx 5.236 \][/tex]
So, the complete solution indicates that the value of [tex]\(x\)[/tex] in the interval [tex]\(\pi \leq x \leq 2\pi\)[/tex] that satisfies the equation [tex]\(\tan^2(x) + 2\sqrt{3}\tan(x) + 3 = 0\)[/tex] is [tex]\( x = \frac{5\pi}{3} \approx 5.236 \)[/tex].
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
