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

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Solution

${sin}^{2} x + 4^{x} - 2^{x + 1} + 1 = 0$
$\Rightarrow (sin x)^{2} + (2^{x})^{2} - {2.2}^{x} + 1 = 0$
$\Rightarrow (sin x)^{2} + (2^{x} - 1)^{2} = 0$
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 $1$ .

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