lgbasallote
Find all values of x such that \(\sin 2x = \sin x\) and 0 < x < 2\(\pi\)
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lgbasallote
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i suppose the only way for this to happen is if the x were 0....
JamesWolf
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yeah or pi, but you specified x < 2pi
lgbasallote
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hmm i suppose pi works
lgbasallote
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sin 2x = sin x
arcsin both sides
2x = x
2x - x = 0
x = 0
how can i get the other values?
JamesWolf
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hmmm good question
leedomathgeek
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sin(2x)=2sin(x)cos(x)
Therefore,
sin(2x)=sin(x) -->
2sin(x)cos(x)=sin(x)
2sin(x)cos(x)-sin(x)=0
sin(x)[2cos(x)-1]=0
Now either
sin(x)=0 ---> x=0,pi,...
or
2cos(x)-1=0 --> cos(x)=1/2 ---> x=pi/3
lgbasallote
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then i suppose the others can be solved by adding pi?
leedomathgeek
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Why do you want to add a pi?
lgbasallote
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because one angle is missing
leedomathgeek
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You have only 3 solutions
leedomathgeek
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0, pi/3, pi
lgbasallote
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there's actually 5.. but i know how to get one of the missing angles...
leedomathgeek
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oh ya i though x between 0 and pi
leedomathgeek
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sure 2pi is the 4th solution
lgbasallote
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yes
leedomathgeek
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and x=2pi-p/3=5pi/3
leedomathgeek
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is the 5th solution
lgbasallote
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oh subtract pi/3 from 2pi...ah yes... the negative angle
leedomathgeek
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exactly ...since cos(x) is positive in the 4th quad
apple_pi
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Here's an example you might want to look at:
tan(4x) = -tan(2x) 0° ≤ x ≤ 360°
tan(4x) = tan(-2x)
****DONT DO THIS: 4x = -2x
INSTEAD... 4x = 180n -2x (Write one side as a general solution)
then simplify:
6x = 180n
x = 30n
then you have all possible values of x
now find all the ones that work, and you have all solutions