\( e^x \sin x = 1 \) 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,
\[
e^x \cos x = -1
\]
If $\alpha$ and $\beta$ are the roots of the equation $2x^{2}+ 2px + p^{2} = 0$, where $p$ is a non-zero real number, and $\alpha^{4}$ and $\beta^{4}$ are the roots of $x^{2} - rx + s = 0$, then the roots of $2x^{2} - 4p^{2}x + 4p^{4} - 2r = 0$ are: