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    hey im a bit stuck on this
    im not 100% sure what to do

    i need to
    solve for values of theta from 0 to 360
    theata = A to make it easier

    sin2AcosA + sin^2A = 1

    what do i do?
    i multipled out to get
    2sinAcosA . cosA + 0.5(1 - cos^A-1)


    also can i get 2sinAcos^2A is that right?

    am i goin right here
    but i dont know where i need to go?

    please help

    thx in advance
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    2sin2AcosA + sin^2 A = 1

    2sinAcos^2A = 1 - sin^2A

    What can we do with the RHS ?

    Heres a spoiler to what you have to do next

    Spoiler:
    Show
    Take the RHS over to the left hand side and factorise. DO NOT DIVIDE you will lose a solution!!!
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    "2sinAcos^2A = 1 - sin^2A"
    if i take it to the lhs i get
    2sinAcos^2A - 1 + sin^2A = 0

    how do i factorise this?

    thx
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    What IS 1-sin^2A though?

    I put the sin^2A over the RHS for a reason!
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    did u mean 1 - sinsquared A
    or 1-sin to power of 2?
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    \large \tex sin^2A = (sinA)^2

    \large \tex 1 - sin^2A = cos^2A

    so it looks like now

    \large \tex 2sinAcos^2A = cos^2A

    can you finish it now?
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    yep - sorry didnt remem that identity i have cos^2 A as something else
    thx man
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    np.
 
 
 
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