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anonymous
 one year ago
Find all solutions to the equation in the interval [0, 2π).
sin 2x  sin 4x = 0
anonymous
 one year ago
Find all solutions to the equation in the interval [0, 2π). sin 2x  sin 4x = 0

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freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin(4x) \neq \sin(2x)+\sin(2x)\]

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.0You need to use \(\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin(2x)\sin(4x) =0 \\ \text{ use double angle identity on the } \sin(4x) \\ \sin(2 \cdot [2x])=2 \sin(2x)\cos(2x) \\ \sin(2x)2 \sin(2x) \cos(2x) =0 \\ \text{ factor out } \sin(2x) \\ \sin(2x)[12\cos(2x)]=0 \\ \text{ set both factors equal to 0} \\ \sin(2x)=0 \text{ or } 12\cos(2x)=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait, 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.0on which part are you confused?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the answer is zero?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0that is only one solution

freckles
 one year ago
Best ResponseYou've already chosen the best response.0can I ask you what you are confused on/

freckles
 one year ago
Best ResponseYou've already chosen the best response.0my work or solving the equations I asked you to solve?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so I started by making an equation 2x4x=0

freckles
 one year ago
Best ResponseYou've already chosen the best response.0where does that equation come from?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0is 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?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0sin(x) is only approximately x for values of x near 0 you cannot use this approximation to solve your equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i was confused by your work

freckles
 one year ago
Best ResponseYou've already chosen the best response.0ok do you know the double angle identities? like for example what does sin(2u)=?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0sin(2u)=2sin(u)cos(u) so if u=2x then we have \[\sin(2[2x])=2\sin(2x)\cos(2x)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[2 \cdot 2 =4 \\ \text{ so } \sin(4x)=\sin(2[2x])=2\sin(2x)\cos(2x) \\ \text{ so I replaced } \sin(4x) \text{ with } 2\sin(2x)\cos(2x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh ok so the answer would be B?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin(2x)\sin(4x)=0 \\ \sin(2x)2 \sin(2x) \cos(2x)=0 \\ \text{ next I noticed a common factor in both terms on the left } \\ \color{red}{\sin(2x)}2 \color{red}{\sin(2x)}\cos(2x)=0\] 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 \[\color{red}{\sin(2x)}[12 \cos(2x)]=0 \\ \text{ so we have to solve both of the following equations } \\ \sin(2x)=0 \text{ or } 12\cos(2x)=0\] 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 \[\sin(u)=0 \text{ or } 12 \cos(u)=0 \\ \sin(u) =0 \text{ or } \cos(u)=\frac{1}{2}\] on the unit circle when do you have the ycoordinates are 0? on the unit circle when do you have the xcoordinates are 1/2?
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