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anonymous

  • one year ago

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

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  1. freckles
    • one year ago
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    \[\sin(4x) \neq \sin(2x)+\sin(2x)\]

  2. ChillOut
    • one year ago
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    You need to use \(\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\)

  3. freckles
    • one year ago
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    \[\sin(2x)-\sin(4x) =0 \\ \text{ use double angle identity on the } \sin(4x) \\ \sin(2 \cdot [2x])=2 \sin(2x)\cos(2x) \\ \sin(2x)-2 \sin(2x) \cos(2x) =0 \\ \text{ factor out } \sin(2x) \\ \sin(2x)[1-2\cos(2x)]=0 \\ \text{ set both factors equal to 0} \\ \sin(2x)=0 \text{ or } 1-2\cos(2x)=0\]

  4. freckles
    • one year ago
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    solve both equations

  5. anonymous
    • one year ago
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    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

  6. freckles
    • one year ago
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    on which part are you confused?

  7. anonymous
    • one year ago
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    so the answer is zero?

  8. freckles
    • one year ago
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    that is only one solution

  9. freckles
    • one year ago
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    there are others

  10. freckles
    • one year ago
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    can I ask you what you are confused on/

  11. freckles
    • one year ago
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    my work or solving the equations I asked you to solve?

  12. anonymous
    • one year ago
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    ok, so I started by making an equation 2x-4x=0

  13. freckles
    • one year ago
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    where does that equation come from?

  14. anonymous
    • one year ago
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    i made sinx=x

  15. freckles
    • one year ago
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    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?

  16. freckles
    • one year ago
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    sin(x) is only approximately x for values of x near 0 you cannot use this approximation to solve your equation

  17. anonymous
    • one year ago
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    i was confused by your work

  18. freckles
    • one year ago
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    ok do you know the double angle identities? like for example what does sin(2u)=?

  19. anonymous
    • one year ago
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    2sin u cos u

  20. freckles
    • one year ago
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    sin(2u)=2sin(u)cos(u) so if u=2x then we have \[\sin(2[2x])=2\sin(2x)\cos(2x)\]

  21. freckles
    • one year ago
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    \[2 \cdot 2 =4 \\ \text{ so } \sin(4x)=\sin(2[2x])=2\sin(2x)\cos(2x) \\ \text{ so I replaced } \sin(4x) \text{ with } 2\sin(2x)\cos(2x)\]

  22. anonymous
    • one year ago
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    oh ok so the answer would be B?

  23. freckles
    • one year ago
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    \[\sin(2x)-\sin(4x)=0 \\ \sin(2x)-2 \sin(2x) \cos(2x)=0 \\ \text{ next I noticed a common factor in both terms on the left } \\ \color{red}{\sin(2x)}-2 \color{red}{\sin(2x)}\cos(2x)=0\] 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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