Solve for x: $sin^2x - 4 sin x + 3 = 0, 0 ≤ x ≤ \frac{\pi}{2}$

$\frac{\pi}{2}$ $\frac{\pi}{4}$ $\frac{\pi}{6}$ $\frac{\pi}{3}$

$\frac{\pi}{2}$

sin²x - 4 sinx + 3 = 0 sin²x - 3 sinx  - sinx + 3 = 0 sinx ( sinx - 3 ) - 1 ( sinx - 3 ) = 0 ( sinx - 1 ) ( sinx - 3 ) = 0 Either sinx - 1 = 0 or sinx - 3 = 0 Sinx - 3 = 0 is not possible So, sinx - 1 = 0 sinx = 1 { we know, sin 90º = 1 } So, x = 90º = \(\frac{π}{2}\)