The number of solutions to the equation $\left|2e^{\sin x}-3-e^{2\sin x}\right|=\left|e^{\sin x}-e^{2\sin x}-1\right|$ in $\left[\mathrm{0,2}\pi \right]$ is

• A

  $0$
• Correct answer
  B

  $2$
• C

  $4$
• D

  $3$

Solution

Putting $e^{\sin x}=t,$ we get, 
$\left|2t-3-t^2\right|=\left|t-t^2-1\right|$
$\left|t^2-2t+3\right|=\left|t^2-t+1\right|$
$\left\{t^2-2t+3>0, \forall t\in R \mathrm{a}\mathrm{n}\mathrm{d} t^2-t+1>0, \forall t\in R \mathrm{a}\mathrm{s} \mathrm{t}\mathrm{h}\mathrm{e}\mathrm{i}\mathrm{r} \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{n}\mathrm{t}<0\right\}$
$\Rightarrow \cancel{t^2}-2t+3=\cancel{t^2}-t+1\Rightarrow t=2$
$\Rightarrow e^{\sin x}=2\Rightarrow \sin x=\ln 2\in \left(\mathrm{0,1}\right)$

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

Hence, the correct answer is option: 'B'.