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Solve trigonometry problem

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data

    Solve ##2sin3x.sinx=1##


    2. Relevant equations


    3. The attempt at a solution
    I used the identity ##( cos (A+B)= cos A cos B- sin A sin B),
    (cos(A-B) = cos A cos B+sinA sinB)→
    -(cos 4x-cos2x)= 2sin 3xsinx, (cos 2x-cos4x=1)##


    now i am stuck , is this correct_
    using ##(cos 2x=2cos^2x-1)⇒(cos 4x=2cos^22x-1),→(2cos^22x-2cos^2x+1=0)##

    is this correct, how do i move folks
     
  2. jcsd
  3. Dec 4, 2016 #2

    cnh1995

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    I'd suggest you use the triple angle formula for sine. It will get much simpler.
    Edit: No I think your current approach is good. Solve the quadratic equation using the discriminant method.
     
  4. Dec 4, 2016 #3

    ehild

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    Substitute 2cos2(2x)-1 for cos(4x) into the equation cos(2x)-cos(4x)=1, and solve for cos(2x) .
     
  5. Dec 4, 2016 #4
    Are you suggesting i use ##sin 3x≡3sin x- 4sin^3x##
     
  6. Dec 4, 2016 #5
    ehild hahahahahahahha that is what i was looking for lol, greetings from Africa bro
     
  7. Dec 4, 2016 #6

    cnh1995

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    No.. I edited my previous post. I was talking about the approach in #3, but it looks like I misread your quadratic equation.
     
  8. Dec 4, 2016 #7
    in that case i just have to solve ## cos 2x(2cos 2x-1)=0## thanks folks
     
  9. Dec 4, 2016 #8
    what do you mean by saying solve the quadratic using discriminant method....
     
  10. Dec 4, 2016 #9

    ehild

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    grandma. :) You are welcome.
     
  11. Dec 4, 2016 #10

    cnh1995

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  12. Dec 5, 2016 #11

    ehild

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    The name is "quadratic formula" . Discriminant method is something different.
     
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