Question

2 cos(x + п/3)= ≥ √2

Answer 1

Ответ:

$2 \cos(x + \frac{\pi}{3} ) \geqslant \sqrt{2} \\ \cos(x + \frac{\pi}{3} ) \geqslant \frac{ \sqrt{2} }{2}$

рисунок

$x + \frac{\pi}{3} ∈[- \frac{\pi}{4} + 2\pi \: n ;\frac{\pi}{4} + 2 \pi \: n] \\ x∈[- \frac{7\pi}{12} + 2\pi \: n ;- \frac{\pi}{12} + 2\pi \: n]$

Answer 2

To solve the inequality $2\cos(x + \pi/3) \geq \sqrt{2}$, you can follow these steps:

1. Start by isolating the cosine term:

   $2\cos(x + \pi/3) \geq \sqrt{2}$

   Divide both sides by 2:

   $\cos(x + \pi/3) \geq \frac{\sqrt{2}}{2}$

2. Next, you need to determine the values of $x$ that satisfy this inequality. You can do this by considering the values of $\cos$ for which it is greater than or equal to $\frac{\sqrt{2}}{2}$.

   Recall that $\cos(\pi/4) = \frac{\sqrt{2}}{2}$.

   So, your inequality becomes:

   $\cos(x + \pi/3) \geq \cos(\pi/4)$

3. Now, we need to find the values of $x$ that make $\cos(x + \pi/3)$ greater than or equal to $\cos(\pi/4)$. To do this, we'll consider the intervals where $\cos(x + \pi/3)$ is greater than or equal to $\cos(\pi/4)$.

   The general form for the cosine function is:

   $\cos(\theta) = \cos(-\theta)$

   This means that $\cos$ is an even function, and its values repeat every $2\pi$ radians. So, you can add or subtract multiples of $2\pi$ to $x$ without changing the inequality.

   $\cos(x + \pi/3) \geq \cos(\pi/4)$ is equivalent to:

   $\cos(x + \pi/3) \geq \cos(\pi/4 + 2k\pi)$, where $k$ is an integer.

4. Now, you can solve for $x + \pi/3$ by taking the arccosine of both sides:

   $x + \pi/3 \geq \pi/4 + 2k\pi$

5. Finally, isolate $x$ by subtracting $\pi/3$ from both sides:

   $x \geq \pi/4 + 2k\pi - \pi/3$

So, the solution to the inequality $2\cos(x + \pi/3) \geq \sqrt{2}$ is:

$x \geq \frac{\pi}{4} + 2k\pi - \frac{\pi}{3}$

Where $k$ is an integer that can take any value (positive or negative) to cover all possible solutions. This inequality represents an infinite number of solutions for $x$.