Discussion
How many solutions are there for the equation \(\cos x - e^{-x} = 0\) ?
Answer: C
The equation has countably infinitely many solutions.
Let \(g(x) = \cos x - e^{-x}\). For large positive \(x\), \(e^{-x}\) is tiny, so \(g(2k\pi) = 1 - e^{-2k\pi} \gt 0\) and \(g((2k + 1)\pi) = -1 - e^{-(2k+1)\pi} \lt 0\). By the intermediate value theorem there is a root in every interval \((2k\pi, (2k + 1)\pi)\), so there are infinitely many roots.
The roots are countable because \(g\) is a non-constant real-analytic function, whose zeros are isolated, and an isolated set in R is countable. (There is also the root \(x = 0\), and none for \(x \lt 0\), where \(e^{-x} \gt 1\).) So 'no solution', 'finitely many' and 'uncountably many' are all wrong.
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