√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 \( k \in \mathbb{Z} \).