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amoodarya

  • one year ago

how do I solve this equation ?

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  1. amoodarya
    • one year ago
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    \[3\sin^3x+2\cos^3x=2sinx+cosx\]

  2. mathslover
    • one year ago
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    I'll be more than fascinated to see what you've tried yet, amoodarya.

  3. amoodarya
    • one year ago
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    I need to solve it analytical ! not by graph , I know one of roots is \[\frac{\pi}{3}\]

  4. mathslover
    • one year ago
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    I wonder, were you given with that root or did you calculate it on your own?

  5. amoodarya
    • one year ago
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    I test it , but I can't solve it nice !

  6. mathslover
    • one year ago
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    One method is to follow the conventional way - that is putting \(\sin x = \sqrt{1- \cos^2 x} \) Although, this is complicated. But, are you studying topic related to systems of equations (or similar to this) ...?

  7. mathslover
    • one year ago
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    And if one of the roots is \(\pi/3 \) then : \(3(3\sqrt{3} /8 ) + 2 (1/8) = 2(\qrt{3})/2 + 1/2 \\ 9\sqrt{3} /8 + 1/4 = \sqrt{3} + 1/2 \) This root doesn't seem to satisfy the given equation.

  8. anonymous
    • one year ago
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    At first sight, I can't see a nice solution to this problem, maybe Weierstrass substitution will work

  9. amoodarya
    • one year ago
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    i solve it numerically , then I saw one root is very close to pi/3 , but missed to check it in equation . thanks for your notification " mukushla"

  10. anonymous
    • one year ago
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    no problem amoodarya

  11. ganeshie8
    • one year ago
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    is this you @amoodarya http://math.stackexchange.com/questions/1344917/how-do-i-solve-this-equation-mathbbr-3-sin3x2-cos3x-2-sin-x-cos

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