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    Supposed to solve cos^2x = 2sin2x (in degrees from 0 -360)

    So I did:
    cos^2x = 4sinxcosx
    cosx = 4sinx
    tanx = 1/4

    x = 14, 194

    But textbook says 90 and 270 are also answers? How?
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    (Original post by G.Y)
    Supposed to solve cos^2x = 2sin2x (in degrees from 0 -360)

    So I did:
    cos^2x = 4sinxcosx
    cosx = 4sinx
    tanx = 1/4

    x = 14, 194

    But textbook says 90 and 270 are also answers? How?
    You cannot divide through by cosx unless you are sure that cosx never equals zero. Instead, you should have moved everything to one side and factorised.
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    (Original post by G.Y)
    Supposed to solve cos^2x = 2sin2x (in degrees from 0 -360)

    So I did:
    cos^2x = 4sinxcosx
    cosx = 4sinx
    tanx = 1/4

    x = 14, 194

    But textbook says 90 and 270 are also answers? How?
    Because thats when \cos x = 0 which is a perfectly valid condition for the original equality to hold.
 
 
 
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