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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].
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