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shaqadry

  • 2 years ago

solve sin^2 x +2 sinx cosx = 0

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  1. beknazar23
    • 2 years ago
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    sinx(sinx+2cosx) = 0 What can you do now? think.

  2. beknazar23
    • 2 years ago
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    if a*b = 0, Either a = 0 or b = 0 Here, a = sinx and b = sinx + 2cosx

  3. satellite73
    • 2 years ago
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    i don't think this is that easy

  4. satellite73
    • 2 years ago
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    that is \(\sin(x)=0\) is easy enough to solve, you get \(x=n\pi, n\in \mathbb{Z}\) but i can't see a good way to solve \[\sin(x)=-2\cos(x)\]

  5. TheViper
    • 2 years ago
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    \[sin^2x+2\ sin\ x\ cos\ x=0\] \[sin\ x(sin\ x+2\ cos\ x)=0\] \[(sin\ x+2\ cos\ x)=0\]

  6. satellite73
    • 2 years ago
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    that is not to say you cannot do it, but it is going to take some work

  7. satellite73
    • 2 years ago
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    the first thing you have to do to solve \[\sin(x)+2\cos(x)=0\] is to write it as a single function of sine

  8. satellite73
    • 2 years ago
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    in general you can write \[a\sin(x)+b\cos(x)\] as \[\sqrt{a^2+b^2}\sin(x+\theta)\]

  9. tayluvsya
    • 2 years ago
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    :'( i need help on mine

  10. ParthKohli
    • 2 years ago
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    I kinda wonder if you could do \(\sin^2(x) + \sin(2x)=0\).

  11. satellite73
    • 2 years ago
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    in this case you get \[\sin(x)+2\cos(x)=\sqrt{5}\sin(x+\theta)\] where \(\tan(\theta)=2\)

  12. beknazar23
    • 2 years ago
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    tanx = -2 x = atan(-2) + n*(pi)

  13. satellite73
    • 2 years ago
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    therefore \(x+\theta=n\pi\) gives \(\theta=n\pi-\tan^{-1}(2)\)

  14. satellite73
    • 2 years ago
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    you can check that it is right by looking here http://www.wolframalpha.com/input/?i=sin%28x%29%2B2cos%28x%29

  15. satellite73
    • 2 years ago
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    @ParthKohli i don't think that identity is going to help in this case you will have two arguments, one of \(x\) and the other of \(2x\) if you have a solution from that approach i would love to see it

  16. TheViper
    • 2 years ago
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    oh I had forgotten this site Thanx @satellite73 for remembering :)

  17. ParthKohli
    • 2 years ago
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    Gaahh @satellite73 you were right.

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