1. **State the problem:** Solve the equation $$\Delta an(2x - \pi) = \frac{\sqrt{2}}{5}$$ for $$x$$. 2. **Recall the definition and properties:** The function $$\Delta an$$ is the hyperbolic tangent function, denoted as $$\tanh$$. So the equation is $$\tanh(2x - \pi) = \frac{\sqrt{2}}{5}$$. 3. **Use the inverse hyperbolic tangent formula:** To solve for $$2x - \pi$$, apply $$\tanh^{-1}$$ to both sides: $$ 2x - \pi = \tanh^{-1}\left(\frac{\sqrt{2}}{5}\right) $$ 4. **Recall the formula for inverse hyperbolic tangent:** $$ \tanh^{-1}(y) = \frac{1}{2} \ln\left(\frac{1+y}{1-y}\right) $$ 5. **Calculate the value:** $$ \tanh^{-1}\left(\frac{\sqrt{2}}{5}\right) = \frac{1}{2} \ln\left(\frac{1 + \frac{\sqrt{2}}{5}}{1 - \frac{\sqrt{2}}{5}}\right) $$ 6. **Simplify the fraction inside the logarithm:** $$ \frac{1 + \frac{\sqrt{2}}{5}}{1 - \frac{\sqrt{2}}{5}} = \frac{\frac{5 + \sqrt{2}}{5}}{\frac{5 - \sqrt{2}}{5}} = \frac{5 + \sqrt{2}}{5 - \sqrt{2}} $$ 7. **Rationalize the denominator:** $$ \frac{5 + \sqrt{2}}{5 - \sqrt{2}} \times \frac{5 + \sqrt{2}}{5 + \sqrt{2}} = \frac{(5 + \sqrt{2})^2}{25 - 2} = \frac{25 + 10\sqrt{2} + 2}{23} = \frac{27 + 10\sqrt{2}}{23} $$ 8. **Substitute back:** $$ 2x - \pi = \frac{1}{2} \ln\left(\frac{27 + 10\sqrt{2}}{23}\right) $$ 9. **Solve for $$x$$:** $$ 2x = \pi + \frac{1}{2} \ln\left(\frac{27 + 10\sqrt{2}}{23}\right) $$ $$ x = \frac{\pi}{2} + \frac{1}{4} \ln\left(\frac{27 + 10\sqrt{2}}{23}\right) $$ 10. **Include the general solution with periodicity:** Since $$\tanh$$ is not periodic, the solution is unique for real $$x$$. However, if the problem context involves periodicity (e.g., from the original trigonometric context), add $$+ k\pi$$ for integer $$k$$ if needed. **Final answer:** $$ x = \frac{\pi}{2} + \frac{1}{4} \ln\left(\frac{27 + 10\sqrt{2}}{23}\right) $$