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Size:  345.7 KB how do I solve for sin4x-sin3x=0 for 0<x<pi? I don't get how the formula sin p- sinq can be used because cos 3.5x will be created... I don't know what to do with the 3.5x ...

    Thanks
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    (Original post by coconut64)
    how do I solve for sin4x-sin3x=0 for 0<x<pi? I don't get how the formula sin p- sinq can be used because cos 3.5x will be created... I don't know what to do with the 3.5x ...

    Thanks
    You don't need to do anything with the 3.5x.


    You have 2 cos 3.5x sin 0.5x = 0

    so, cos 3.5x = 0 or sin 0.5x = 0

    If sin 0.5x = 0, then 0.5x = 0, pi, 2pi, 3pi, etc.
    Hence x = 0, 2pi, 4pi, 6pi, etc.
    Now restrict it to the desired domain, x = 0 (everything else is outside the given domain)

    Now do the same with the cos. It's slightly trickier with the values, but the process is the same.
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    Alternatively you could solve \sin4\theta = \sin 3\theta by considering the sine graph / CAST.

    This is quicker but can be confusing for a lot of students.

    E.g. the first solution is when 4\theta = 3\theta
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    (Original post by notnek)
    Alternatively you could solve \sin4\theta = \sin 3\theta by considering the sine graph / CAST.

    This is quicker but can be confusing for a lot of students.

    E.g. the first solution is when 4\theta = 3\theta
    I don't think I had ever learnt that method to solve it. I do know the cast diagram but not the method but thanks
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    (Original post by ghostwalker)
    You don't need to do anything with the 3.5x.


    You have 2 cos 3.5x sin 0.5x = 0

    so, cos 3.5x = 0 or sin 0.5x = 0

    If sin 0.5x = 0, then 0.5x = 0, pi, 2pi, 3pi, etc.
    Hence x = 0, 2pi, 4pi, 6pi, etc.
    Now restrict it to the desired domain, x = 0, or 2pi.

    Now do the same with the cos. It's slightly trickier with the values, but the process is the same.
    Thanks I managed to get it now.
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    (Original post by coconut64)
    Thanks I managed to get it now.
    Note: I've amended my previous post, as the 2pi solution is outside the given domain and so not valid.
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    (Original post by coconut64)
    I don't think I had ever learnt that method to solve it. I do know the cast diagram but not the method but thanks
    If you want to have a go then I recommend you watch the solution to 5c here.

    That video shows how you can solve

    \displaystyle \tan\left(\theta + \frac{\pi}{6}\right) = \tan\left(\pi-\theta\right)

    which is similar to your problem.
 
 
 
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