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