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    2sin2x+cos2x=1 solve by r formula

    I used r formula
    R sin(2x+a)

    I got
    rsina =1
    rcos1=2
    R= sqrt(5)
    a= 26.57 degrees

    now solve:
    so sqrt(5) sin (2x+26.57)=1
    sin(2x+26.57)=1/sqrt(5)
    2x+26.57= arcsin (1/sqrt(5))

    arcsin (1/sqrt(5)) = 26.57 degrees

    so 2x =0 so x=0 so angles are 0 and 180 degrees

    I got that part right!

    But it says the angles are also 63.43 and 243.43 degrees! HOW??!
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    (Original post by 0utdoorz)
    2sin2x+cos2x=1 solve by r formula

    I used r formula
    R sin(2x+a)

    I got
    rsina =1
    rcos1=2
    R= sqrt(5)
    a= 26.57 degrees

    now solve:
    so sqrt(5) sin (2x+26.57)=1
    sin(2x+26.57)=1/sqrt(5)
    2x+26.57= arcsin (1/sqrt(5))

    arcsin (1/sqrt(5)) = 26.57 degrees

    so 2x =0 so x=0 so angles are 0 and 180 degrees

    I got that part right!

    But it says the angles are also 63.43 and 243.43 degrees! HOW??!
    By CAST, you should know that there are an infinite number of solutions to sine and cosine equations. e.g. sin30 = 0.5 But sin 150 = 0.5 also. And sin 390 = 0.5 (To see this, draw a sine graph)

    So arcsin (1/sqrt(5)) = 26.6. But it also equals 180-26.6 = 153.4 and it also equals 360 + 26.6 = 386.6 degrees. (This is fairly standard CAST).
    (the question should specify a range of degrees that you should consider).

    When you rearrange to find x in these 3 solutions, you should find the required answer.
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    Assuming your range is from 0 < x < 360

    New range: 26.57 < (2x + 26.57) < 746.57 hence solutions are found when:

    2x + 26.57 = 26.57

    2x + 26.57 = 180 - (26.57)

    2x + 26.57 = 360 + (26.57)

    2x + 26.57 = 540 - (26.57)
 
 
 
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Updated: May 16, 2013
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