+0  
 
0
3264
3
avatar

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

Guest Apr 18, 2015

Best Answer 

 #3
avatar+93616 
+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
avatar+93616 
+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+27032 
+5

x = pi is also a solution

.

Alan  Apr 18, 2015
 #3
avatar+93616 
+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

32 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.