pt lượg giác đề thi thử đêy!

N

nhocngo976

[tex]{\frac{sin^3.sin3x+cos^3.cos3x}{tan(x-{\frac{\pi}{6}}).tan(x+{\frac{\pi}{3})}}=-{\frac{1}{8}[/tex]
post bài giải cho mìk đi

ĐK...

mẫu số: [TEX]tan(x- \frac{\pi}{6}).tan(x+ \frac{\pi}{3})=\frac{sin(x+ \frac{\pi}{3}- \frac{\pi}{2})}{cos(x+ \frac{\pi}{3}- \frac{\pi}{2})}. \frac{sin(x+\frac{\pi}{3})}{cos(x+\frac{\pi}{3})}= - \frac{cos(x+\frac{\pi}{3})}{sin(x+\frac{\pi}{3})}.\frac{sin(x+\frac{\pi}{3})}{cos(x+\frac{\pi}{3})}=-1[/TEX]

\Rightarrowpt \Leftrightarrow[TEX]sin3x.\frac{3sinx-sin3x}{4}+\frac{cos3x+3cosx}{4}.cos3x=\frac{1}{8}[/TEX]

\Leftrightarrow[TEX]6sinxsin3x+6cosxcos3x+2cos^23x-2sin^23x =1[/TEX]

\Leftrightarrow[TEX]6cos2x+2cos6x-1=0[/TEX]

\Leftrightarrow[TEX]8cos^32x-6cos2x+6cos2x-1=0[/TEX]

\Leftrightarrow[TEX]8cos^32x-1=0[/TEX]
 
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N

nhoc_maruko9x

ĐK...

mẫu số: [TEX]tan(x- \frac{\pi}{6}).tan(x+ \frac{\pi}{3})=\frac{sin(x+ \frac{\pi}{3}- \frac{\pi}{2})}{cos(x+ \frac{\pi}{3}- \frac{\pi}{2})}. \frac{sin(x+\frac{\pi}{3})}{cos(x+\frac{\pi}{3})}= - \frac{cos(x+\frac{\pi}{3})}{sin(x+\frac{\pi}{3})}.\frac{sin(x+\frac{\pi}{3})}{cos(x+\frac{\pi}{3})}=-1[/TEX]

\Rightarrowpt \Leftrightarrow[TEX]sin3x.\frac{3sinx-sin3x}{4}+\frac{cos3x+3cosx}{4}.cos3x=\frac{1}{8}[/TEX]

\Leftrightarrow[TEX]3sinxsin3x+3cosxcos3x+cos^23x-sin^23x =0[/TEX]

\Leftrightarrow[TEX]3cos2x+cos6x=0[/TEX]

\Leftrightarrow[TEX]4cos^32x-3cos2x+3cos2x=0[/TEX]

\Leftrightarrow[TEX]cos2x=0[/TEX]
Nhầm lẫn chỗ nào rồi.. cos2x = 0 không là nghiệm của PT :)
 
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