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

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

Guest Apr 18, 2015

#3
+91412
+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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#1
+91412
+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
+26397
+5

x = pi is also a solution

.

Alan  Apr 18, 2015
#3
+91412
+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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