Question

The number of solution(s) of the equation ${\sin }^{2}x+4^x-2^{x+1}+1=0$ is

Answer 1

${\sin }^{2}x+4^x-2^{x+1}+1=0$
$\Rightarrow (\sin x{)}^2+(2^x{)}^2-{2.2}^x+1=0$
$\Rightarrow (\sin x{)}^2+(2^x-1{)}^2=0$
This is only possible when
$\Rightarrow \sin x=0$ and $2^x-1=0$
$\Rightarrow \sin x=0$ and $2^x=1$
$\Rightarrow x=0$

Hence, the number of solution is $1$.