1. **State the problem:** Solve the equation $$e^{-x} - \cos x = 0$$ for $x$. 2. **Rewrite the equation:** We want to find $x$ such that $$e^{-x} = \cos x.$$ 3. **Important notes:** - The function $e^{-x}$ is always positive and decreases as $x$ increases. - The function $\cos x$ oscillates between $-1$ and $1$. 4. **Check for possible solutions:** Since $e^{-x} > 0$, solutions must satisfy $\cos x > 0$, which happens near $x=0$ and periodically. 5. **Test $x=0$:** $$e^{0} = 1, \quad \cos 0 = 1,$$ so $x=0$ is a solution. 6. **Check behavior near $x=0$:** - For small positive $x$, $e^{-x}$ decreases from 1, $\cos x$ decreases from 1. - For $x$ near 1, $e^{-1} \approx 0.3679$, $\cos 1 \approx 0.5403$, so $e^{-x} < \cos x$. 7. **Graphical or numerical methods:** The equation is transcendental; exact algebraic solutions are not possible. 8. **Summary:** The only exact solution is $x=0$. Other solutions can be approximated numerically if needed. **Final answer:** $$x = 0.$$