IDNLearn.com makes it easy to get reliable answers from experts and enthusiasts alike. Our platform is designed to provide quick and accurate answers to any questions you may have.
Sagot :
To solve the equation [tex]\(\cos \left(\frac{x}{2}\right) = \frac{\sqrt{3}}{2}\)[/tex], we need to determine the values of [tex]\(x\)[/tex] for which this equation holds true. Let's follow the steps to find the solutions:
1. Identify the standard angles:
First, we need to determine the angles [tex]\(\theta\)[/tex] at which [tex]\(\cos(\theta) = \frac{\sqrt{3}}{2}\)[/tex].
We know from trigonometry that [tex]\(\cos(\theta) = \frac{\sqrt{3}}{2}\)[/tex] at [tex]\(\theta = \frac{\pi}{6}\)[/tex] and [tex]\(\theta = -\frac{\pi}{6}\)[/tex].
2. Relate the standard angles to the given equation:
We are given [tex]\(\cos \left(\frac{x}{2}\right) = \frac{\sqrt{3}}{2}\)[/tex]. This means that:
[tex]\[ \frac{x}{2} = \frac{\pi}{6} \quad \text{or} \quad \frac{x}{2} = -\frac{\pi}{6} \][/tex]
3. Solve for [tex]\(x\)[/tex]:
Multiply both sides of each equation by 2 to find [tex]\(x\)[/tex]:
[tex]\[ x = 2 \left(\frac{\pi}{6}\right) = \frac{2\pi}{6} = \frac{\pi}{3} \][/tex]
[tex]\[ x = 2 \left(-\frac{\pi}{6}\right) = -\frac{2\pi}{6} = -\frac{\pi}{3} \][/tex]
4. Verify the periodicity:
The cosine function is periodic with a period of [tex]\(2\pi\)[/tex]. While [tex]\(\frac{\pi}{3}\)[/tex] and [tex]\(-\frac{\pi}{3}\)[/tex] are primary solutions, all solutions can be expressed in terms of the general solution for [tex]\(\cos(\theta) = \frac{\sqrt{3}}{2}\)[/tex]:
[tex]\[ x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = -\frac{\pi}{3} + 2k\pi \quad \text{for integer } k \][/tex]
Since the initial problem did not specify additional periods, the solutions within one period [tex]\(0 \leq x < 2\pi\)[/tex] are:
[tex]\[ x_1 \approx 1.047 \text{ (radians) and }, \quad x_2 \approx -1.047 \text{ (radians)} \][/tex]
Thus, the detailed, step-by-step solution results in:
[tex]\[ (1.0471975511965979, -1.0471975511965979) \][/tex]
These are the values of [tex]\(x\)[/tex] that satisfy the given equation [tex]\(\cos \left(\frac{x}{2}\right) = \frac{\sqrt{3}}{2}\)[/tex].
1. Identify the standard angles:
First, we need to determine the angles [tex]\(\theta\)[/tex] at which [tex]\(\cos(\theta) = \frac{\sqrt{3}}{2}\)[/tex].
We know from trigonometry that [tex]\(\cos(\theta) = \frac{\sqrt{3}}{2}\)[/tex] at [tex]\(\theta = \frac{\pi}{6}\)[/tex] and [tex]\(\theta = -\frac{\pi}{6}\)[/tex].
2. Relate the standard angles to the given equation:
We are given [tex]\(\cos \left(\frac{x}{2}\right) = \frac{\sqrt{3}}{2}\)[/tex]. This means that:
[tex]\[ \frac{x}{2} = \frac{\pi}{6} \quad \text{or} \quad \frac{x}{2} = -\frac{\pi}{6} \][/tex]
3. Solve for [tex]\(x\)[/tex]:
Multiply both sides of each equation by 2 to find [tex]\(x\)[/tex]:
[tex]\[ x = 2 \left(\frac{\pi}{6}\right) = \frac{2\pi}{6} = \frac{\pi}{3} \][/tex]
[tex]\[ x = 2 \left(-\frac{\pi}{6}\right) = -\frac{2\pi}{6} = -\frac{\pi}{3} \][/tex]
4. Verify the periodicity:
The cosine function is periodic with a period of [tex]\(2\pi\)[/tex]. While [tex]\(\frac{\pi}{3}\)[/tex] and [tex]\(-\frac{\pi}{3}\)[/tex] are primary solutions, all solutions can be expressed in terms of the general solution for [tex]\(\cos(\theta) = \frac{\sqrt{3}}{2}\)[/tex]:
[tex]\[ x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = -\frac{\pi}{3} + 2k\pi \quad \text{for integer } k \][/tex]
Since the initial problem did not specify additional periods, the solutions within one period [tex]\(0 \leq x < 2\pi\)[/tex] are:
[tex]\[ x_1 \approx 1.047 \text{ (radians) and }, \quad x_2 \approx -1.047 \text{ (radians)} \][/tex]
Thus, the detailed, step-by-step solution results in:
[tex]\[ (1.0471975511965979, -1.0471975511965979) \][/tex]
These are the values of [tex]\(x\)[/tex] that satisfy the given equation [tex]\(\cos \left(\frac{x}{2}\right) = \frac{\sqrt{3}}{2}\)[/tex].
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for visiting IDNLearn.com. We’re here to provide accurate and reliable answers, so visit us again soon.