The problem asks us to solve several trigonometric equations and inequalities within specified intervals. Specifically, we need to solve the following: * $cos(x) = \frac{1}{2}$ for $x \in \mathbb{R}$ * $sin(2x - \frac{\pi}{3}) = sin(\frac{\pi}{4} - x)$ for $x \in [0, 2\pi]$ * $cos(2x) = \frac{\sqrt{3}}{2}$ for $x \in ]-\pi, \pi[$ * $sin(x) \ge \frac{1}{2}$ for $x \in ]-\pi, \pi]$ * $cos(x) < \frac{1}{2}$ for $x \in [0, 2\pi]$

1. Problem Description

The problem asks us to solve several trigonometric equations and inequalities within specified intervals. Specifically, we need to solve the following:

* $cos(x) = \frac{1}{2}$ for $x \in \mathbb{R}$

* $sin(2x - \frac{\pi}{3}) = sin(\frac{\pi}{4} - x)$ for $x \in [0, 2\pi]$

* $cos(2x) = \frac{\sqrt{3}}{2}$ for $x \in ]-\pi, \pi[$

* $sin(x) \ge \frac{1}{2}$ for $x \in ]-\pi, \pi]$

* $cos(x) < \frac{1}{2}$ for $x \in [0, 2\pi]$

2. Solution Steps

* Equation 1: $cos(x) = \frac{1}{2}$ for $x \in \mathbb{R}$

The general solution for $cos(x) = \frac{1}{2}$ is $x = \pm \frac{\pi}{3} + 2k\pi$, where $k \in \mathbb{Z}$.

* Equation 2: $sin(2x - \frac{\pi}{3}) = sin(\frac{\pi}{4} - x)$ for $x \in [0, 2\pi]$

We know that if $sin(a) = sin(b)$, then $a = b + 2k\pi$ or $a = \pi - b + 2k\pi$, where $k \in \mathbb{Z}$.

Case 1: $2x - \frac{\pi}{3} = \frac{\pi}{4} - x + 2k\pi$

$3x = \frac{\pi}{4} + \frac{\pi}{3} + 2k\pi = \frac{7\pi}{12} + 2k\pi$

$x = \frac{7\pi}{36} + \frac{2k\pi}{3}$

For $k = 0$, $x = \frac{7\pi}{36}$.

For $k = 1$, $x = \frac{7\pi}{36} + \frac{2\pi}{3} = \frac{7\pi + 24\pi}{36} = \frac{31\pi}{36}$.

For $k = 2$, $x = \frac{7\pi}{36} + \frac{4\pi}{3} = \frac{7\pi + 48\pi}{36} = \frac{55\pi}{36}$.

For $k = 3$, $x = \frac{7\pi}{36} + \frac{6\pi}{3} = \frac{7\pi + 72\pi}{36} = \frac{79\pi}{36} > 2\pi$, so we stop here.

Case 2: $2x - \frac{\pi}{3} = \pi - (\frac{\pi}{4} - x) + 2k\pi$

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

$x = \pi - \frac{\pi}{4} + \frac{\pi}{3} + 2k\pi = \frac{12\pi - 3\pi + 4\pi}{12} + 2k\pi = \frac{13\pi}{12} + 2k\pi$

For $k = 0$, $x = \frac{13\pi}{12}$.

For $k = 1$, $x = \frac{13\pi}{12} + 2\pi = \frac{13\pi + 24\pi}{12} = \frac{37\pi}{12} > 2\pi$, so we stop here.

Therefore, the solutions are $x = \frac{7\pi}{36}, \frac{31\pi}{36}, \frac{55\pi}{36}, \frac{13\pi}{12}$.

* Equation 3: $cos(2x) = \frac{\sqrt{3}}{2}$ for $x \in ]-\pi, \pi[$

Since $cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, we have $2x = \pm \frac{\pi}{6} + 2k\pi$.

$x = \pm \frac{\pi}{12} + k\pi$

For $k = 0$, $x = \pm \frac{\pi}{12}$.

For $k = 1$, $x = \frac{\pi}{12} + \pi = \frac{13\pi}{12}$ and $x = -\frac{\pi}{12} + \pi = \frac{11\pi}{12}$.

For $k = -1$, $x = \frac{\pi}{12} - \pi = -\frac{11\pi}{12}$ and $x = -\frac{\pi}{12} - \pi = -\frac{13\pi}{12}$.

Since $x \in ]-\pi, \pi[$, the solutions are $x = -\frac{11\pi}{12}, -\frac{\pi}{12}, \frac{\pi}{12}, \frac{11\pi}{12}$.

* Inequality 1: $sin(x) \ge \frac{1}{2}$ for $x \in ]-\pi, \pi]$

We know that $sin(x) = \frac{1}{2}$ when $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. In the interval $[0, 2\pi]$, $sin(x) \ge \frac{1}{2}$ for $x \in [\frac{\pi}{6}, \frac{5\pi}{6}]$. Since we are considering the interval $]-\pi, \pi]$, we also need to consider the sine values in the negative x direction. Since $sin(-x) = -sin(x)$, $sin(x) > 0$ when $x$ is between $0$ and $\pi$. $sin(x) \ge \frac{1}{2}$ in the interval $[\frac{\pi}{6}, \frac{5\pi}{6}]$. Therefore, the solution is $x \in [\frac{\pi}{6}, \frac{5\pi}{6}]$.

* Inequality 2: $cos(x) < \frac{1}{2}$ for $x \in [0, 2\pi]$

We know that $cos(x) = \frac{1}{2}$ when $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$.

Since $cos(x)$ is decreasing on $[0, \pi]$ and increasing on $[\pi, 2\pi]$, the solutions are $x \in (\frac{\pi}{3}, \frac{5\pi}{3})$.

3. Final Answer

* $cos(x) = \frac{1}{2}$: $x = \pm \frac{\pi}{3} + 2k\pi$, where $k \in \mathbb{Z}$.

* $sin(2x - \frac{\pi}{3}) = sin(\frac{\pi}{4} - x)$: $x = \frac{7\pi}{36}, \frac{31\pi}{36}, \frac{55\pi}{36}, \frac{13\pi}{12}$.

* $cos(2x) = \frac{\sqrt{3}}{2}$: $x = -\frac{11\pi}{12}, -\frac{\pi}{12}, \frac{\pi}{12}, \frac{11\pi}{12}$.

* $sin(x) \ge \frac{1}{2}$: $x \in [\frac{\pi}{6}, \frac{5\pi}{6}]$.

* $cos(x) < \frac{1}{2}$: $x \in (\frac{\pi}{3}, \frac{5\pi}{3})$.