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    solve the following equation for -Pi<0<Pi
    sin4theta = cos 2theta

    I tried sin(2theta+2theta) but still

    Please provide full working.
    Thx
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    I reckon this'll work; I'll be using x instead of theta as it's more concise.

    sin(4x)=cos(2x)
    expand LHS using sin double angle formula
    2sin(2x)cos(2x)=cos(2x)
    divide both sides by cos(2x)
    2sin(2x)=1
    divide by two
    sin(2x)=1/2

    can you take it from here? (hint: I think there are four solutions in the range...)

    Note: I think this is right, but haven't done this for a while now...
    If it is wrong the error will be from dividing by cos(2x) - do reply and let me know if this is or isn't valid.

    Hope this helps!
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    (Original post by alan.accordion)
    I reckon this'll work; I'll be using x instead of theta as it's more concise.

    sin(4x)=cos(2x)
    expand LHS using sin double angle formula
    2sin(2x)cos(2x)=cos(2x)
    divide both sides by cos(2x)
    2sin(2x)=1
    divide by two
    sin(2x)=1/2

    can you take it from here? (hint: I think there are four solutions in the range...)

    Note: I think this is right, but haven't done this for a while now...
    You can't (or shouldn't) divide by cos2x here. It's part of the answer you're looking for, so by dividing through by it, you're elminating (potentially) 2 solutions.

    I would keep extrapolating by using the double angle formulae. It will reduce, eventually, to an equation containing only one circle function.
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    (Original post by alan.accordion)
    I reckon this'll work; I'll be using x instead of theta as it's more concise.

    sin(4x)=cos(2x)
    expand LHS using sin double angle formula
    2sin(2x)cos(2x)=cos(2x)
    Up to here is right, but remember cos(2x) could be equal to 0 so you can't divide through by it.
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    (Original post by mufas)
    solve the following equation for -Pi<0<Pi
    sin4theta = cos 2theta

    I tried sin(2theta+2theta) but still

    Please provide full working.
    Thx
    did this in 30 secs :P. Not the solution but the method. Sin4A= 2sin(2A). Cos (2A)

    now sin 4A = cos 2A, thus: 2sin 2A.cos2A-Cos2A = 0

    Factorising: Cos 2A*(2Sin2A -1) = 0.

    You solve this between the limits.
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    (Original post by blue_shift86)
    did this in 30 secs :P. Not the solution but the method. Sin4A= 2sin(2A). Cos (2A)

    now sin 4A = cos 2A, thus: 2sin 2A.cos2A-Cos2A = 0

    Factorising: Cos 2A*(2Sin2A -1) = 0.

    You solve this between the limits.
    huh :confused:? Didn't get it
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    (Original post by mufas)
    huh :confused:? Didn't get it
    Essentially, he's saying you can take the Cos2x over to the other side, expand Sin4x (Sin(2x+2x)using the double angle identity, and take out Cos2x as a factor.

    It's a legitimate way to do it. You just have to remember that each trig function you're doing is in the interval 0<x<4pi, so you're going to have a fair few solutions, (i.e, go through 2 periods of the respective trig graphs, or around the unit circle twice).
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    Hint: note that \cos{2\theta} = \sin\left(2\theta+\frac{\pi}{2} \right) and recall that if \sin{\lambda} = \sin{\alpha} then \lambda = 2n\pi+\alpha or (2n+1)\pi-\alpha.
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    (Original post by mufas)
    huh :confused:? Didn't get it
    Apologies, let me try again:

    Q) sin (4x) = cos (2x)

    Sin (A+B) = SinA. CosB + SinB. Cos A. [Sin two angle identity)
    If you let A and B both be 2x, then you get:

    LHS: Sin (2x+2x) = Sin(2x).Cos(2x) + Sin(2x).Cos(2x) = 2Sin(2x).Cos(2x)
    RHS = cos (2x)

    Now LHS=RHS

    2Sin(2x). Cos(2x) = Cos (2x) [take cos 2x to other side)
    2Sin(2x).Cos(2x) - Cos (2x) = 0
    Now factorise out Cos (2x)

    So: Cos(2x).[2Sin(2x) - 1] = 0

    Now you just solve it normally between -pi and +pi.

    I hope this helps. Sorry for being a bit late. I've been gardening for the last few hours
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    (Original post by Piecewise)
    Hint: note that \cos{2\theta} = \sin\left(2\theta+\frac{\pi}{2} \right) and recall that if \sin{\lambda} = \sin{\alpha} then \lambda = 2n\pi+\alpha or (2n+1)\pi-\alpha.
    Give over! If the opening poster couldn't follow the solutions given earlier in the thread, they'll have no chance with that.
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    (Original post by blue_shift86)
    Apologies, let me try again:

    Q) sin (4x) = cos (2x)

    Sin (A+B) = SinA. CosB + SinB. Cos A. [Sin two angle identity)
    If you let A and B both be 2x, then you get:

    LHS: Sin (2x+2x) = Sin(2x).Cos(2x) + Sin(2x).Cos(2x) = 2Sin(2x).Cos(2x)
    RHS = cos (2x)

    Now LHS=RHS

    2Sin(2x). Cos(2x) = Cos (2x) [take cos 2x to other side)
    2Sin(2x).Cos(2x) - Cos (2x) = 0
    Now factorise out Cos (2x)

    So: Cos(2x).[2Sin(2x) - 1] = 0

    Now you just solve it normally between -pi and +pi.

    I hope this helps. Sorry for being a bit late. I've been gardening for the last few hours
    Thank you soo much!!! :eek:
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    (Original post by alan.accordion)
    I reckon this'll work; I'll be using x instead of theta as it's more concise.

    sin(4x)=cos(2x)
    expand LHS using sin double angle formula
    2sin(2x)cos(2x)=cos(2x)
    divide both sides by cos(2x)
    2sin(2x)=1
    divide by two
    sin(2x)=1/2

    can you take it from here? (hint: I think there are four solutions in the range...)

    Note: I think this is right, but haven't done this for a while now...
    If it is wrong the error will be from dividing by cos(2x) - do reply and let me know if this is or isn't valid.

    Hope this helps!
    thank you There are 8 answers between -pi<0<pi :eek:
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    (Original post by mufas)
    thank you There are 8 answers between -pi<0<pi :eek:
    no problem! that's why i hate questions like this, lol! Too many sinver cos and sin crap to do! :P and too much +ing and -ing :P
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    Hmm, instead of that division of cos(2x) earlier, you could just factorise it and go from there if I'm not mistaken.
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    (Original post by Piecewise)
    Hint: note that \cos{2\theta} = \sin\left(2\theta+\frac{\pi}{2} \right) and recall that if \sin{\lambda} = \sin{\alpha} then \lambda = 2n\pi+\alpha or (2n+1)\pi-\alpha.
    yer yer, quit showing off now douche bag.
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    Jeez, was my hint really that hard?
 
 
 

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