Between any two real roots of the equation $e^x sin x = 1$, the equation $e^x cos x = –1$ has
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Number of Roots
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,
\[
e^x \cos x = -1
\]