Question

2cos(x-П/6)>=корень из 3

Answer 1

Cos(x-п/6)=корень3/2
Х-п/6=+ - арксоsкорня3/2 +2пн где н принадлежитz
X-п/6=+-п/6+2пн
Х=+-п/6+п/6+2пн
Х первая=п/6+п/6+2пн=п/3+2пн
Или
Х вторая=-п/6+п/6+2пн=2пн

Answer 2

I can help you with math problems, among other things. You asked me to answer in detail: 2cos(x-П/6)>=корень из 3. Here is what I found:

The first step is to divide both sides of the inequality by 2, which gives us:

$$\cos(x-\frac{\pi}{6}) \geq \frac{\sqrt{3}}{2}$$

Next, we need to find the values of x that satisfy this inequality. We can use the inverse cosine function, or arccos, to find the angle whose cosine is equal to $\frac{\sqrt{3}}{2}$. This angle is $\frac{\pi}{6}$ radians, or 30 degrees. However, there are infinitely many angles that have the same cosine value, because the cosine function is periodic and repeats every $2\pi$ radians, or 360 degrees. Therefore, we need to add or subtract multiples of $2\pi$ to the angle $\frac{\pi}{6}$ to get all the possible solutions. We can write this as:

$$x - \frac{\pi}{6} = \arccos(\frac{\sqrt{3}}{2}) \pm 2\pi n$$

where n is any integer. To isolate x, we add $\frac{\pi}{6}$ to both sides, which gives us:

$$x = \frac{\pi}{6} + \arccos(\frac{\sqrt{3}}{2}) \pm 2\pi n$$

Since $\arccos(\frac{\sqrt{3}}{2}) = \frac{\pi}{6}$, we can simplify this further to:

$$x = \frac{\pi}{3} \pm 2\pi n$$

This means that x can be any angle that is equal to $\frac{\pi}{3}$ radians, or 60 degrees, plus or minus a multiple of $2\pi$ radians, or 360 degrees. We can write this in interval notation as:

$$x \in [\frac{\pi}{3} \pm 2\pi n, \frac{\pi}{3} \pm 2\pi (n+1)]$$

where n is any integer. This is the final answer.

If you want to see the graph of the function $y = 2\cos(x-\frac{\pi}{6})$, you can use this [online calculator](https://www.mathway.com/ru/popular-problems/Trigonometry/374925). You can also check your answer by plugging in different values of x and seeing if they satisfy the inequality.

I hope this helps you understand how to solve this problem. If you have any other questions, feel free to ask me.