√2sin x-1=0

To solve the equation \( \sqrt{2} \sin x - 1 = 0 \), start by isolating \( \sin x \):

\[ \sqrt{2} \sin x = 1 \]

Now, divide both sides by \( \sqrt{2} \):

\[ \sin x = \frac{1}{\sqrt{2}} \]

We can simplify \( \frac{1}{\sqrt{2}} \) to \( \frac{\sqrt{2}}{2} \). So, we have:

\[ \sin x = \frac{\sqrt{2}}{2} \]

The sine function equals \( \frac{\sqrt{2}}{2} \) at specific angles. The general solutions for \( \sin x = \frac{\sqrt{2}}{2} \) are:

\[ x = \frac{\pi}{4} + 2k\pi \quad \text{and} \quad x = \frac{3\pi}{4} + 2k\pi \]

where \( k \) is any integer (representing the periodic nature of the sine function).

Thus, the complete solution is:

\[ x = \frac{\pi}{4} + 2k\pi \quad \text{and} \quad x = \frac{3\pi}{4} + 2k\pi \]

where \( k \in \mathbb{Z} \).