Explanation
The slope(s) of common tangent(s) to the curves $$ \displaystyle y={ e }^{ -x }$$ and $$ \displaystyle y={ e }^{ -x }\sin { x } $$ can be -
$$ \displaystyle y={ e }^{ -x }\ \ \&\ y={ e }^{ -x }\sin { x } $$
$$ \displaystyle { y }^{ ' }=-{ e }^{ -x }...(i)\quad \& \quad { y }^{ ' }=-{ e }^{ -x }\left( \sin { x } -\cos { x } \right) ...(ii)$$
equating (i) and (ii)
$$ \Rightarrow \displaystyle { e }^{ -x }\left( 1-\sin { x } +\cos { x } \right) =0$$
$$ \displaystyle { e }^{ -x }\neq 0\quad \Rightarrow \quad 1-\sin { x } +\cos { x } =0$$
$$ \displaystyle \Rightarrow \quad 2\cos ^{ 2 }{ \frac { x }{ 2 } } \quad =\quad 2\sin { \frac { x }{ 2 } } \cos { \frac { x }{ 2 } } $$
$$ \displaystyle \Rightarrow \quad 2\cos { \frac { x }{ 2 } } \left( \sin { \frac { x }{ 2 } -\cos { \frac { x }{ 2 } } } \right) =0 \Rightarrow x=\frac { \pi }{ 2 } ,\pi $$
Slope can be $$ \displaystyle -{ e }^{ -\tfrac{\pi}2 }$$ and $$ \displaystyle -{ e }^{ -\pi }$$
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