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    The question is to solve the equation 2cos2x +sqrt5 sin2x =3 between -π and π
    I used the equation Rsin(2x+α) I found R to be 3 and α to be 0.841
    I'm having trouble solving this equation, not sure whether it is beacuse of the R and α values.

    Thank you in advance!!!
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    Your alpha is a bit out.
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    (Original post by Mr M)
    Your alpha is a bit out.
    how will I work out alpha? I've tried working it out but not what I've done wrong :/
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    (Original post by Lavanyaa)
    how will I work out alpha? I've tried working it out but not what I've done wrong :/
    Expand 3 \sin(2x+\alpha) ?
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    (Original post by Mr M)
    Expand 3 \sin(2x+\alpha) ?
    As the equation is equal to 3 don't you need to take the 3 to the other side and divide it by the other 3 ?
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    (Original post by Lavanyaa)
    As the equation is equal to 3 don't you need to take the 3 to the other side and divide it by the other 3 ?
    You need to find alpha first.

    Do you actually understand how to express a number in harmonic form?
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    (Original post by Mr M)
    You need to find alpha first.

    Do you actually understand how to express a number in harmonic form?
    I wasn't taught this way, is it easier?
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    (Original post by Lavanyaa)
    I wasn't taught this way, is it easier?
    Easier than what? Show your working.

    The only alternative is making a half tangent substitution. That is a little quicker but I'm fairly sure you are supposed to be using the line of attack you proposed in your question.
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    (Original post by Mr M)
    Easier than what? Show your working.

    The only alternative is making a half tangent substitution. That is a little quicker but I'm fairly sure you are supposed to be using the line of attack you proposed in your question.
    I found α to be 0.7297...
    so one of the values will be 0.4205
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    (Original post by Lavanyaa)
    I found α to be 0.7297...
    so one of the values will be 0.4205
    yes
 
 
 
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