Thread: 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..

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,









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

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

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