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HelpMe94

  • 2 years ago

Sin(2x)*Sin(x)=Cos(x) ..... Solve X

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  1. HelpMe94
    • 2 years ago
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    Sin(2x)= 2Sin(x)Cos(x)

  2. HelpMe94
    • 2 years ago
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    So I Got ..... (2Sin(x)Cos(x)Sin(x))-Cos(x)=0

  3. HelpMe94
    • 2 years ago
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    Now What?

  4. mark_o.
    • 2 years ago
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    try factoring them (2Sin(x)Cos(x)Sin(x))-Cos(x)=0

  5. HelpMe94
    • 2 years ago
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    How?

  6. HelpMe94
    • 2 years ago
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    @mark_o. How?

  7. HelpMe94
    • 2 years ago
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    Cos(x) (2 Sin(x) Sin(x))-1=0

  8. HelpMe94
    • 2 years ago
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    Cos(x) (2Sin(x)Sin(x))=1

  9. mark_o.
    • 2 years ago
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    hm lets try this (2Sin(x)Cos(x)Sin(x)) = Cos(x) divide both sides by cos x then (2Sin(x)Sin(x)) = 1 can you continue from here?

  10. HelpMe94
    • 2 years ago
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    @mark_o. is it possible to combine the sines?

  11. mark_o.
    • 2 years ago
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    yes

  12. HelpMe94
    • 2 years ago
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    How?

  13. mark_o.
    • 2 years ago
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    (2Sin(x)Sin(x)) = 1 multiplying 2[Sin(x)]^2 = 1

  14. HelpMe94
    • 2 years ago
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    oh i didn't see that 2 was not part of sine

  15. mark_o.
    • 2 years ago
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    ok from here 2[Sin(x)]^2 = 1 [Sin(x)]^2 = 1/2

  16. HelpMe94
    • 2 years ago
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    ok then Sin(x)^2 = 1-Cos(2x)/2 Right?

  17. mark_o.
    • 2 years ago
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    nope, use power of i/2 on both sides from [Sin(x)]^2 = 1/2 ([Sin(x)]^2)^1/2 = sqrt(1/2) correct? yes or no?

  18. HelpMe94
    • 2 years ago
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    How would this make it solve so i get answers to the unit circle?

  19. mark_o.
    • 2 years ago
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    ([Sin(x)]^2)^1/2 = sqrt(1/2) \[\sin x=\sqrt{\frac{ 1 }{ 2 }}=\frac{ 1 }{ \sqrt{2} }*\frac{ \sqrt{2} }{ \sqrt{2} }=...?\]

  20. HelpMe94
    • 2 years ago
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    X=pi/4

  21. mark_o.
    • 2 years ago
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    yes...you are correct x=45deg =pi/4

  22. HelpMe94
    • 2 years ago
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    although this seems not right there should be more to this.... because i thought.... Sin(2x)Sin(x)=Cos(x) Sin(2x)Sin(x)-Cos(x)=0 2Sin(x)Cos(x)Sin(x)-Cos(x)=0 Cos(x)(2Sin(x)Sin(x))=1 Then Maybe..... Cos(x)(2Sin(x)Sin(x))=1 Cos(x)=1 and 2Sin(x)=1 and Sin(x)=1

  23. HelpMe94
    • 2 years ago
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    Attached is a example of this..

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  24. HelpMe94
    • 2 years ago
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    if it is then Cos(x)=1 has solutions 0 and 2pi 2Sin(x)=1 also as Sin(x)=1/2 solutions pi/6 and 5pi/6 Sin(x)=1 solution pi/2

  25. mark_o.
    • 2 years ago
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    yesssssss thats right 2Sin(x)Cos(x)Sin(x)-Cos(x)=0 could be like Cos(x)[(2Sin(x)Sin(x))-1]=0 cos x=0 and 2Sin(x)Sin(x))-1=0 x= arc cos 0 x=90deg=pi/2 and for 2Sin(x)Sin(x))-1]=0 2Sin(x)Sin(x))=1

  26. HelpMe94
    • 2 years ago
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    as long as restrictions are [0,2pi)

  27. mark_o.
    • 2 years ago
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    yes you got it.. lol..:D

  28. HelpMe94
    • 2 years ago
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    oh that makes so much more sense after the factoring and combining of sines... lol

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