# Thread: Simple double angle formulae

1. ## Simple double angle formulae

Hi,

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,

Solve 2sin^2x = sin2x, 0<x<pi

Thanks!

p.s I know its simple, but I'm stuck for some reason..
Hi CSG18,

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

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

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

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

$\displaystyle 2\sin x=0 \:\r\:\: \sin x - \cos x=0$

$\displaystyle \sin x=0 \:\r\:\: \sin x = \cos x$

$\displaystyle \boxed{x=0} \:\r \:\:\boxed{x=\frac{\pi}{4}}$