The number of solutions to the equation |2e^ sin x - 3 - e^ 2sin x...

Putting $e^{sin x} = t,$ we get,&
\left|2t - 3 - t^{2}\right| =\left|t - t^{2} - 1\right|
\left|t^{2} - 2t + 3\right| =\left|t^{2} - t + 1\right|
\left\{t^{2} - 2t + 3 > 0, \forall t \in Randt^{2} - t + 1 > 0, \forall t \in Rastheirdiscriminant < 0\right\}
$⇒ t^{2} - 2t + 3 = t^{2} - t + 1 ⇒ t = 2$
$⇒ e^{sin x} = 2 ⇒ sin x = \ln 2 \in \left(0,1\right)$

Hence, from the graph, we get, in $\left[0,2\pi\right]$& there are two solutions