Between any two real roots of the equation $e^x sin x = 1$, the equation $e^x cos x = –1$ has

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, \[ e^x \cos x = -1 \]

\[ \boxed{\text{At least one root}} \]