You are given the equation:

$\sin(\sqrt{2}x) - \cos\left(\frac{x}{2}\right) = 0$

To solve this, we want to find values of $x$ that satisfy the equation. Let's follow these steps:

Step 1: Isolate one of the trigonometric functions

Rearrange the equation:

$\sin(\sqrt{2}x) = \cos\left(\frac{x}{2}\right)$

Step 2: Use a trigonometric identity

Recall the identity $\cos(\theta) = \sin\left(\frac{\pi}{2} - \theta\right)$. So, we can rewrite the equation as:

$\sin(\sqrt{2}x) = \sin\left(\frac{\pi}{2} - \frac{x}{2}\right)$

Step 3: Solve the sine equation

For $\sin A = \sin B$, the general solution is:

$A = B + 2k\pi \quad \text{or} \quad A = \pi - B + 2k\pi \quad \text{for} \quad k \in \mathbb{Z}$

Thus, we have two cases:

1. $\sqrt{2}x = \frac{\pi}{2} - \frac{x}{2} + 2k\pi$
2. $\sqrt{2}x = \pi - \left(\frac{\pi}{2} - \frac{x}{2}\right) + 2k\pi$

Step 4: Solve each case

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

Multiply through by 2 to eliminate the fraction:

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

Now, solve for $x$:

$2\sqrt{2}x + x = \pi + 4k\pi$

Factor out $x$:

$x(2\sqrt{2} + 1) = \pi(1 + 4k)$

Thus,

$x = \frac{\pi(1 + 4k)}{2\sqrt{2} + 1}$

Case 2: $\sqrt{2}x = \pi - \left(\frac{\pi}{2} - \frac{x}{2}\right) + 2k\pi$

Simplify the right-hand side:

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

$\sqrt{2}x = \frac{\pi}{2} + \frac{x}{2} + 2k\pi$

Multiply through by 2 to eliminate the fraction:

$2\sqrt{2}x = \pi + x + 4k\pi$

Now, solve for $x$:

$2\sqrt{2}x - x = \pi(1 + 4k)$

Factor out $x$:

$x(2\sqrt{2} - 1) = \pi(1 + 4k)$

Thus,

$x = \frac{\pi(1 + 4k)}{2\sqrt{2} - 1}$

Final Answer

The solutions to the equation are:

$x = \frac{\pi(1 + 4k)}{2\sqrt{2} + 1} \quad \text{or} \quad x = \frac{\pi(1 + 4k)}{2\sqrt{2} - 1}, \quad k \in \mathbb{Z}$

Would you like further clarification on any of the steps?