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..

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- Apr 20th 2010, 08:49 AMCSG18Simple 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.. - Apr 20th 2010, 10:21 AMmasters
Hi CSG18,

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

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

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

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

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

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

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