Find all solutions to the equation in the interval [0, 2π). sin 2x - sin 4x = 0

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Find all solutions to the equation in the interval [0, 2π). sin 2x - sin 4x = 0

Mathematics
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\[\sin(4x) \neq \sin(2x)+\sin(2x)\]
You need to use \(\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\)
\[\sin(2x)-\sin(4x) =0 \\ \text{ use double angle identity on the } \sin(4x) \\ \sin(2 \cdot [2x])=2 \sin(2x)\cos(2x) \\ \sin(2x)-2 \sin(2x) \cos(2x) =0 \\ \text{ factor out } \sin(2x) \\ \sin(2x)[1-2\cos(2x)]=0 \\ \text{ set both factors equal to 0} \\ \sin(2x)=0 \text{ or } 1-2\cos(2x)=0\]

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Other answers:

solve both equations
wait, I'm so confused. here are the options i have A) π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6 B) 0, π/6, π/2, 5π/6, π, 7π/6, 3π/2, 11π/6 C) 0, 2π/3, 4π/3 D) 0, π/3, 2π/3, π, 4π/3, 5π/3
on which part are you confused?
so the answer is zero?
that is only one solution
there are others
can I ask you what you are confused on/
my work or solving the equations I asked you to solve?
ok, so I started by making an equation 2x-4x=0
where does that equation come from?
i made sinx=x
is there anyway you can tell me what you are confused on? was it my work or me asking you to solve those two equations I asked you to solve?
sin(x) is only approximately x for values of x near 0 you cannot use this approximation to solve your equation
i was confused by your work
ok do you know the double angle identities? like for example what does sin(2u)=?
2sin u cos u
sin(2u)=2sin(u)cos(u) so if u=2x then we have \[\sin(2[2x])=2\sin(2x)\cos(2x)\]
\[2 \cdot 2 =4 \\ \text{ so } \sin(4x)=\sin(2[2x])=2\sin(2x)\cos(2x) \\ \text{ so I replaced } \sin(4x) \text{ with } 2\sin(2x)\cos(2x)\]
oh ok so the answer would be B?
\[\sin(2x)-\sin(4x)=0 \\ \sin(2x)-2 \sin(2x) \cos(2x)=0 \\ \text{ next I noticed a common factor in both terms on the left } \\ \color{red}{\sin(2x)}-2 \color{red}{\sin(2x)}\cos(2x)=0\] if you want to put a 1 next to sin(2x) you can you can do this since 1*sin(2x) is sin(2x) \[1 \cdot \color{red}{\sin(2x)}-2 \color{red}{\sin(2x)}\cos(2x)=0\] now factor out the thing in red \[\color{red}{\sin(2x)}[1-2 \cos(2x)]=0 \\ \text{ so we have to solve both of the following equations } \\ \sin(2x)=0 \text{ or } 1-2\cos(2x)=0\] if you want to replace 2x with u and solve for u first then you can come back later and solve for x but replacing u back with 2x \[\sin(u)=0 \text{ or } 1-2 \cos(u)=0 \\ \sin(u) =0 \text{ or } \cos(u)=\frac{1}{2}\] on the unit circle when do you have the y-coordinates are 0? on the unit circle when do you have the x-coordinates are 1/2?

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