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

1. Problem Description

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

2. Solution Steps

We need to find the values of $x$ that satisfy the equation $2sin^2(x) - \sqrt{2}cos(x) = 0$.

We can use the trigonometric identity $sin^2(x) + cos^2(x) = 1$ to express $sin^2(x)$ in terms of $cos^2(x)$.

$sin^2(x) = 1 - cos^2(x)$

Substituting this into the equation, we get:

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

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

Rearranging the terms, we get a quadratic equation in terms of $cos(x)$:

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

Let $y = cos(x)$. Then the equation becomes:

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

We can solve for $y$ using the quadratic formula:

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

where $a=2$, $b=\sqrt{2}$, and $c=-2$.

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

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

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

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

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 $cos(x) = \frac{\sqrt{2}}{2}$ or $cos(x) = -\sqrt{2}$.

Since $-1 \leq cos(x) \leq 1$, the second solution $cos(x) = -\sqrt{2}$ is not possible as $-\sqrt{2} \approx -1.414$.

Therefore, $cos(x) = \frac{\sqrt{2}}{2}$.

The values of $x$ for which $cos(x) = \frac{\sqrt{2}}{2}$ are $x = \frac{\pi}{4} + 2n\pi$ and $x = -\frac{\pi}{4} + 2n\pi$, where $n$ is an integer. We can also write $x = \frac{7\pi}{4} + 2n\pi$.

Combining the two solutions gives $x = \pm \frac{\pi}{4} + 2n\pi$, where $n$ is an integer.

3. Final Answer

$x = \frac{\pi}{4} + 2n\pi$ and $x = \frac{7\pi}{4} + 2n\pi$, where $n$ is an integer. Or $x = \pm \frac{\pi}{4} + 2n\pi$, where $n$ is an integer.