Làm hộ mình bài lượng giác với

N

nguyenbahiep1

(sinX)^8+(cosX)^8=2(sinX)^10+2(cosX)^10+cos(2X)
Giải chi tiết cho mình nhé tks


[laTEX](sin^8x+cos^8x)(sin^2x+cos^2x) = 2(sin^{10}x+cos^{10}x) + cos2x \\ \\ sin^{10}x+cos^{10}x - sin^2x.cos^8x - cos^2x.sin^8x + cos2x = 0 \\ \\ cos^8x(cos^2x-sin^2x) + sin^8x(sin^2-cos^2x) + cos2x = 0 \\ \\ cos^8x.cos2x - sin^8x.cos2x + cos2x =0 \\ \\ TH_1: cos2x = 0 \\ \\ TH_2: cos^8x - sin^8x+1 = 0 \\ \\(cos^2x+sin^2x)(cos^2x-sin^2x)(cos^4x+sin^4x) + 1 = 0 \\ \\ cos2x.(1-2sin^2x.cos^2x) + 1 =0 \\ \\ cos2x.(1- \frac{1}{2}sin^22x) + 1 = 0 \\ \\ cos2x( \frac{1}{2}+ \frac{1}{2}cos^22x) + 1 = 0 \\ \\ u + u^3 + 2 = 0 \Rightarrow u = - 1 \Rightarrow cos2x = - 1[/laTEX]
 
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