Hi,
Can someone please show me how to answer this question:
Solve 2sin^2x = sin2x, 0<x<pi
Thanks!
p.s I know its simple, but I'm stuck for some reason..
Hi CSG18,
$\displaystyle 2 \sin^2 x=\sin 2x$
$\displaystyle 2 \sin^2 x = 2 \sin x \cos x$
$\displaystyle 2\sin^2 x-2\sin x \cos x = 0$
$\displaystyle 2\sin x(\sin x - \cos x)=0$
$\displaystyle 2\sin x=0 \:\r\:\: \sin x - \cos x=0$
$\displaystyle \sin x=0 \:\r\:\: \sin x = \cos x$
$\displaystyle \boxed{x=0} \:\r \:\:\boxed{x=\frac{\pi}{4}}$