# Math Help - Simple double angle formulae

1. ## Simple double angle formulae

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..

2. Originally Posted by CSG18
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,

$2 \sin^2 x=\sin 2x$

$2 \sin^2 x = 2 \sin x \cos x$

$2\sin^2 x-2\sin x \cos x = 0$

$2\sin x(\sin x - \cos x)=0$

$2\sin x=0 \:\r\:\: \sin x - \cos x=0" alt="2\sin x=0 \:\r\:\: \sin x - \cos x=0" />

$\sin x=0 \:\r\:\: \sin x = \cos x" alt="\sin x=0 \:\r\:\: \sin x = \cos x" />

$\boxed{x=0} \:\r \:\:\boxed{x=\frac{\pi}{4}}" alt="\boxed{x=0} \:\r \:\:\boxed{x=\frac{\pi}{4}}" />