trigonometryy

sin^4x-cos^4x=7/2(sinxcosx)

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  1. (sin^2x + cos^2x)(sin^2x - cos^2x) = 7/2(sinxcosx)
    1(sin^2x - cos^2x) = 7/2(sinxcosx)
    cos^2x - sin^2x = -(7/2)(1/2)(2sinxcosx)
    cos 2x = -(7/4)sin 2x
    -4/7 = sin2x/cos2x
    tan2x = -4/7

    so 2x is in the 2nd or 4th quadrants
    2x = 150.26° or 2x = 330.255
    x = 75.13° or x = 165.13°

    since the period of tan 2x is 90°, adding 90° to each answer gives us more answers
    e.g. 165.13 + 90 = 255.13° etc

    I had my calculator set to degrees, switch it to radians if you need radian answers

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  2. sin^4x+cos^4x=7/2(sinxcosx)
    if it is with +..??how can u do it..can you solve it plz,cuz i have an exam

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  3. you mean sin^4x+cos^4x=7/2(sinxcosx)

    that would be much messier.

    I graphed the left side as y = sin^4x+cos^4x
    and the right side as y = (7/2)sinxcosx on the same grid and noticed that they intersect at
    x = 15° , 75° , 195‚ and 255° , so there are actually "nice" solutions, I checked them , they work.

    At the moment I don't see an easy way to actually work it out,

    How about something like this

    sin^2x(1-cos^2x) + cos^2x(1-sin^2x) for the left side
    = sin^2x - sin^2xcos^2x + cos^2x - sin^2xcos^2x
    = 1 - 2 sin^2cos^2

    so 1 - 2 sin^2cos^2 = (7/2)sinxcosx
    looks promising ....
    2 - 4sin^2xcos^2x = 7sinxcosx
    but 2sinxcosx = sin 2x
    2 - (sin2x)^2 = (7/2)(2sinxcosx)
    4 - 2(sin2x)^2 = 7sin2x
    2sin 2x + 7sin2x - 4 = 0
    (2sin2x - 1)(sin2x + 4) = 0
    sin2x - 1/2 or sin2x = -4 --- > not possible

    sin2x = 1/2
    2x = 30 or 150
    x = 15 or 75 , period of sin2x is 180, so add 180 to the answers ..... x = 195 , etc

    YEAHHHH

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  4. wow..thnx so much..u really are a genius

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