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Cos70cos10+sin70sin10 e =2cos^2 2x-1

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    Cos70cos10+sin70sin10 e =2cos^2 2x-1
    don't know what that "e" is supposed to be, seems to be superfluous.

    recall : cos (A-B) = cosAcoB + sinAsinB
    Cos70cos10+sin70sin10 =2cos^2 2x-1
    cos(70-10) = 2cos^2 2x - 1
    cos60 = 2cos^2 2x - 1
    recall cos 2A = 2cos^2 A - 1
    so 2cos^2 2x - 1 = cos 4x

    cos 4x = cos60
    4x = 60° or 4x = 300°
    x = 15° or x = 75°

    since cos 2x has a period of 180°
    other solutions would be
    x = 15+180 = 195°
    or
    x = 75+180 = 255°

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