math

sorry but i don't understand this one either
solve 4sin^2x+4√2 cosx-6 for all real values of x

asked by isaiah
  1. don't forget your Algebra I now that you're in trig:

    4sin^2x+4√2 cosx-6
    4(1-cos^2x)+4√2 cosx-6
    4 - 4cos^2x + 4√2 cosx - 6 = 0
    4cos^2x - 4√2 cosx + 2 = 0
    2cos^2x - 2√2 cosx + 1 = 0
    (√2 cosx - 1)^2 = 0
    cosx = 1/√2
    x = 2kπ ± π/4

    that last line because cos π/4 = cos -π/4 = 1/√2

    posted by Steve

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