Thread: solving for x

0 <= x <= 2pi

(sin^2x)-(2sinx)=3

Originally Posted by wisezeta

0 <= x <= 2pi

(sin^2x)-(2sinx)=3

$\displaystyle \sin^2{x} - 2\sin{x} - 3 = 0$

$\displaystyle
(\sin{x}-3)(\sin{x}+1) = 0
$

$\displaystyle
x = \frac{3\pi}{2}
$