The problem is to solve the trigonometric equation $2\sin^2{x} - \sqrt{2}\cos{x} = 0$.

1. Problem Description

The problem is to solve the trigonometric equation $2\sin^2{x} - \sqrt{2}\cos{x} = 0$.

2. Solution Steps

First, we use the identity $\sin^2{x} + \cos^2{x} = 1$ to rewrite $\sin^2{x}$ as $1 - \cos^2{x}$.

Substituting this into the equation, we get:

$2(1 - \cos^2{x}) - \sqrt{2}\cos{x} = 0$

$2 - 2\cos^2{x} - \sqrt{2}\cos{x} = 0$

$-2\cos^2{x} - \sqrt{2}\cos{x} + 2 = 0$

$2\cos^2{x} + \sqrt{2}\cos{x} - 2 = 0$

Let $y = \cos{x}$. Then the equation becomes:

$2y^2 + \sqrt{2}y - 2 = 0$

We can use the quadratic formula to solve for $y$:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 2$, $b = \sqrt{2}$, and $c = -2$. So,

$y = \frac{-\sqrt{2} \pm \sqrt{(\sqrt{2})^2 - 4(2)(-2)}}{2(2)}$

$y = \frac{-\sqrt{2} \pm \sqrt{2 + 16}}{4}$

$y = \frac{-\sqrt{2} \pm \sqrt{18}}{4}$

$y = \frac{-\sqrt{2} \pm 3\sqrt{2}}{4}$

So, we have two possible values for $y$:

$y_1 = \frac{-\sqrt{2} + 3\sqrt{2}}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}$

$y_2 = \frac{-\sqrt{2} - 3\sqrt{2}}{4} = \frac{-4\sqrt{2}}{4} = -\sqrt{2}$

Since $y = \cos{x}$, we have two cases:

Case 1: $\cos{x} = \frac{\sqrt{2}}{2}$

The solutions for this are $x = \frac{\pi}{4} + 2\pi k$ and $x = \frac{7\pi}{4} + 2\pi k$, where $k$ is an integer. These can also be written as $x = \pm \frac{\pi}{4} + 2\pi k$.

Case 2: $\cos{x} = -\sqrt{2}$

Since the range of cosine is $[-1, 1]$, there is no solution for this case because $-\sqrt{2} < -1$.

3. Final Answer

The solutions are $x = \frac{\pi}{4} + 2\pi k$ and $x = \frac{7\pi}{4} + 2\pi k$, where $k$ is an integer. These can also be written as $x = \pm \frac{\pi}{4} + 2\pi k$.