The given equation is:

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

We know that:

$\sin^2(x) + \cos^2(x) = 1$

So the equation simplifies to:

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

Next, solve for $\cos(2x)$:

$\cos(2x) = \frac{1}{\sqrt{2}}$

The cosine of $2x$ equals $\frac{1}{\sqrt{2}}$ at angles where $2x = \frac{\pi}{4} + 2k\pi$ or $2x = \frac{7\pi}{4} + 2k\pi$, where $k$ is an integer.

Thus:

$2x = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad 2x = \frac{7\pi}{4} + 2k\pi$

Now, solve for $x$ by dividing by 2:

$x = \frac{\pi}{8} + k\pi \quad \text{or} \quad x = \frac{7\pi}{8} + k\pi$

Since we need to find the solutions in the interval $(0, 2\pi)$, we consider $k = 0$ and $k = 1$:

1. For $k = 0$:

   • $x = \frac{\pi}{8}$
   • $x = \frac{7\pi}{8}$

2. For $k = 1$:

   • $x = \frac{\pi}{8} + \pi = \frac{9\pi}{8}$
   • $x = \frac{7\pi}{8} + \pi = \frac{15\pi}{8}$

Therefore, the solutions on the interval $(0, 2\pi)$ are:

$x = \frac{\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{15\pi}{8}$

Would you like any details or have any further questions?

Here are some related questions you might find interesting:

1. How would the solutions change if the interval was $[0, 4\pi]$?
2. What is the general solution to the equation $\cos(2x) = \frac{1}{\sqrt{2}}$?
3. How would the problem change if $\sin^2(x) + \cos^2(x)$ was a different constant?
4. What would the solutions be if the equation was $\sqrt{2} \sin(2x) = 1$ instead?
5. How would the equation change if we added a phase shift to $\cos(2x)$?

Tip: Remember that $\cos(2x)$ and $\sin(2x)$ represent periodic functions, so be sure to consider all relevant solutions within the given interval.