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

Guest Apr 18, 2015

Best Answer 

 #3
avatar+90988 
+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
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3+0 Answers

 #1
avatar+90988 
+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
 #2
avatar+26322 
+5

x = pi is also a solution

.

Alan  Apr 18, 2015
 #3
avatar+90988 
+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

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