[Toán 11] Phương trình lượng giác

N

nguyenbahiep1

câu b

[laTEX] -cos4x + sin6x = cos2x \\ \\ cos4x + cos2x = 2sin3x.cos3x \\ \\ cos3x.cosx = sin3x.cos3x \\ \\ TH_1: cos3x = 0 \\ \\ TH_2: cosx = sin 3x[/laTEX]
 
T

thienlong233

cos2x-2cosx+sinx=cosx(cos2-sin2x)
giải

<-->cos2x+2cosx+sinx=cosx(2cos^2(x)-1)-2cos^2(x).sinx
<-->cos2x+2cosx+sinx=2cos^3(x)-cosx-2cos^2(x).sinx
<-->{2cos^2(x)-1+2cosx-2cos^3(x)}+sinx(1+2cos^2(x))=0
<-->-2(cos^(x)-1)(cosx-1)+sinx(1+2cos^2(x))
<-->2sin^2(x).(cosx-1)+sinx(1+2cos^2(x))
<-->sinx(2cosx-2+1+2cos^2(x))=0
giải tiếp bạn nhé :D
 
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