The problem consists of four parts: 1. Verify the equality $\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}$.

1. Problem Description

The problem consists of four parts:

1. Verify the equality $\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}$.

2. Solve the quadratic equation $2x^2 + (1 - \sqrt{2})x - \frac{\sqrt{2}}{2} = 0$ and the inequality $2x^2 + (1 - \sqrt{2})x - \frac{\sqrt{2}}{2} > 0$ in the real numbers.

3. Find the solutions of the trigonometric equation $2\cos^2x + (1 - \sqrt{2})\cos x - \frac{\sqrt{2}}{2} = 0$. Represent the solutions on the unit circle.

4. Find the solutions in $[0, 2\pi]$ of the inequality $2\cos^2x + (1 - \sqrt{2})\cos x - \frac{\sqrt{2}}{2} > 0$. Represent the solutions on the unit circle.

2. Solution Steps

1. Verify the equality:

$(1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}$. Thus $\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}$.

2. Solve the quadratic equation and inequality:

The equation is $2x^2 + (1 - \sqrt{2})x - \frac{\sqrt{2}}{2} = 0$.

We can multiply by 2 to get $4x^2 + 2(1 - \sqrt{2})x - \sqrt{2} = 0$.

The discriminant is $\Delta = (1 - \sqrt{2})^2 - 4(2)(-\frac{\sqrt{2}}{2}) = 1 - 2\sqrt{2} + 2 + 4\sqrt{2} = 3 + 2\sqrt{2} = (1 + \sqrt{2})^2$.

Thus the solutions are $x_{1,2} = \frac{-(1 - \sqrt{2}) \pm \sqrt{(1 + \sqrt{2})^2}}{4} = \frac{-1 + \sqrt{2} \pm (1 + \sqrt{2})}{4}$.

$x_1 = \frac{-1 + \sqrt{2} + 1 + \sqrt{2}}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}$.

$x_2 = \frac{-1 + \sqrt{2} - 1 - \sqrt{2}}{4} = \frac{-2}{4} = -\frac{1}{2}$.

Now, consider the inequality $2x^2 + (1 - \sqrt{2})x - \frac{\sqrt{2}}{2} > 0$.

Since the leading coefficient is positive, the parabola opens upward. The inequality is satisfied when $x < -\frac{1}{2}$ or $x > \frac{\sqrt{2}}{2}$.

3. Solve the trigonometric equation:

Let $y = \cos x$. Then the equation becomes $2y^2 + (1 - \sqrt{2})y - \frac{\sqrt{2}}{2} = 0$.

The solutions are $y_1 = \frac{\sqrt{2}}{2}$ and $y_2 = -\frac{1}{2}$.

So $\cos x = \frac{\sqrt{2}}{2}$ or $\cos x = -\frac{1}{2}$.

If $\cos x = \frac{\sqrt{2}}{2}$, then $x = \pm \frac{\pi}{4} + 2k\pi$, $k \in \mathbb{Z}$.

If $\cos x = -\frac{1}{2}$, then $x = \pm \frac{2\pi}{3} + 2k\pi$, $k \in \mathbb{Z}$.

The solutions are $x = \frac{\pi}{4} + 2k\pi, -\frac{\pi}{4} + 2k\pi, \frac{2\pi}{3} + 2k\pi, -\frac{2\pi}{3} + 2k\pi$ for $k \in \mathbb{Z}$.

4. Solve the trigonometric inequality in $[0, 2\pi]$:

$2\cos^2x + (1 - \sqrt{2})\cos x - \frac{\sqrt{2}}{2} > 0$.

Let $y = \cos x$. Then we have $2y^2 + (1 - \sqrt{2})y - \frac{\sqrt{2}}{2} > 0$. The solutions are $y < -\frac{1}{2}$ or $y > \frac{\sqrt{2}}{2}$.

So $\cos x < -\frac{1}{2}$ or $\cos x > \frac{\sqrt{2}}{2}$.

$\cos x > \frac{\sqrt{2}}{2}$ when $x \in [0, \frac{\pi}{4}) \cup (\frac{7\pi}{4}, 2\pi]$.

$\cos x < -\frac{1}{2}$ when $x \in (\frac{2\pi}{3}, \frac{4\pi}{3})$.

Thus, the solution is $x \in [0, \frac{\pi}{4}) \cup (\frac{2\pi}{3}, \frac{4\pi}{3}) \cup (\frac{7\pi}{4}, 2\pi]$.

3. Final Answer

1. $\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}$

2. The solutions to the equation are $x = \frac{\sqrt{2}}{2}$ and $x = -\frac{1}{2}$. The solution to the inequality is $x < -\frac{1}{2}$ or $x > \frac{\sqrt{2}}{2}$.

3. The solutions to the trigonometric equation are $x = \pm \frac{\pi}{4} + 2k\pi, \pm \frac{2\pi}{3} + 2k\pi$ for $k \in \mathbb{Z}$.

4. The solution to the trigonometric inequality in $[0, 2\pi]$ is $x \in [0, \frac{\pi}{4}) \cup (\frac{2\pi}{3}, \frac{4\pi}{3}) \cup (\frac{7\pi}{4}, 2\pi]$.