Find all values of x such that \(\sin 2x = \sin x\) and 0 < x < 2\(\pi\)

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Find all values of x such that \(\sin 2x = \sin x\) and 0 < x < 2\(\pi\)

Trigonometry
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i suppose the only way for this to happen is if the x were 0....
yeah or pi, but you specified x < 2pi
hmm i suppose pi works

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Other answers:

sin 2x = sin x arcsin both sides 2x = x 2x - x = 0 x = 0 how can i get the other values?
hmmm good question
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
then i suppose the others can be solved by adding pi?
Why do you want to add a pi?
because one angle is missing
You have only 3 solutions
0, pi/3, pi
there's actually 5.. but i know how to get one of the missing angles...
oh ya i though x between 0 and pi
sure 2pi is the 4th solution
yes
and x=2pi-p/3=5pi/3
is the 5th solution
oh subtract pi/3 from 2pi...ah yes... the negative angle
exactly ...since cos(x) is positive in the 4th quad
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

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