Given:
e^x \sin x = 1 has two real roots → say x_1 and x_2
Apply Rolle’s Theorem:
Since f(x) = e^x \sin x is continuous and differentiable, and f(x_1) = f(x_2) , ⇒ There exists c \in (x_1, x_2) such that f'(c) = 0
Compute:
f'(x) = e^x(\sin x + \cos x) = 0 \Rightarrow \tan x = -1 At this point, e^x \cos x = -1
\boxed{\text{At least one root}}
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