#3**+5 **

**Thanks Alan,**

I had to think about that one, but yes of course any x where sin x =0 will be a solution.

**This is why it is best to use facorization to make sure you get ALL the solutions.**

$$\\2sinxcosx=2sinx\\

2sinxcosx-2sinx=0\\

2sinx(cosx-1)=0\\

sinx=0\;\;OR\;\;cosx=1\\

x=0,\pi\;\; OR\;\;x=0\\

$The 2 solution are $x=0\:\;and\;\;x=\pi$$

Melody
Apr 19, 2015

#1**+5 **

how do I solve for x in, sin2x=2sinx when in the intervals of [0,2pi)?

$$\\2sinxcosx=2sinx\\

cosx=1\\

x=0$$

Melody
Apr 18, 2015

#3**+5 **

Best Answer

**Thanks Alan,**

I had to think about that one, but yes of course any x where sin x =0 will be a solution.

**This is why it is best to use facorization to make sure you get ALL the solutions.**

$$\\2sinxcosx=2sinx\\

2sinxcosx-2sinx=0\\

2sinx(cosx-1)=0\\

sinx=0\;\;OR\;\;cosx=1\\

x=0,\pi\;\; OR\;\;x=0\\

$The 2 solution are $x=0\:\;and\;\;x=\pi$$

Melody
Apr 19, 2015