1. **State the problem:** Solve the equation $x^2 - 2y^2 = 1$ for $y$ in terms of $x$. 2. **Rewrite the equation:** The equation is $x^2 - 2y^2 = 1$. 3. **Isolate $y^2$:** $$x^2 - 2y^2 = 1 \implies -2y^2 = 1 - x^2 \implies 2y^2 = x^2 - 1$$ 4. **Divide both sides by 2:** $$y^2 = \frac{x^2 - 1}{2}$$ 5. **Take the square root of both sides:** $$y = \pm \sqrt{\frac{x^2 - 1}{2}}$$ 6. **Interpretation:** The solution expresses $y$ in terms of $x$. For real values of $y$, the expression under the square root must be non-negative, so $x^2 - 1 \geq 0$, or $|x| \geq 1$. **Final answer:** $$y = \pm \sqrt{\frac{x^2 - 1}{2}}$$