Question

the number of real numbers in [0,2π] satisfying sin ^ 4 x - 2 sin^2x+ 1 is​

Answer 1

        $\text{The given equation is }sin^4x - 2sin^2x + 1 = 0$

        $\text{Let }m = sin^2x$

$\implies \: m^2 - 2m + 1 =0$

        $\text{Applying splitting the middle term method}$

$\implies \: m^2 - m - m + 1 =0$

$\implies \: m(m-1)-1(m-1) = 0$

$\implies \: (m-1)^2 = 0$

$\implies \: m = 1$

$\implies \: sin^2x = 1$

$\implies \: sin^2x = sin^2\text{\huge{(}}\dfrac{\pi}{2}\text{\huge{)}}$

$\implies \: x = n\pi \± \dfrac{\pi}{2} \text{ where }n \: \epsilon \: I$

        $\text{It is also given that }x \: \epsilon \: [0,2\pi]$

$\implies \: \boxed{x = \dfrac{\pi}{2},\dfrac{3\pi}{2}}$