[TOÁN 11] giải pt lượng giác

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botvit

1. [TEX]{sin}^{8}x[/TEX]+[TEX]{cos}^{8}x[/TEX]=2([TEX]{sin}^{10}x[/TEX]+[TEX]{cos}^{10}x[/TEX])+[TEX]\frac{5}{4}cos2x[/TEX]

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PT\Leftrightarrow[TEX]sin^8x(2sin^2x-1)+cos^8x(2cos^2x-1)+\frac{5}{4}cos2x=0[/TEX]
\Leftrightarrow[TEX]cos2x(cos^8x-sin^8x+\frac{5}{4})=0[/TEX]
ta co[TEX]cos^8x-sin^8x=(cos^4x-sin^4x)(cosx^4x+sin^4x)=cos2x(1-\frac{1}{2}.sin^22x)=cos2x(1-\frac{1-cos^22x}{2})[/TEX]
thay vao` ok?
 
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