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hoanghondo94

[TEX]sinx +coxsin2x+\sqrt{3}cos3x=2(cos4x +sin^3x)[/TEX]
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Mình giúp bạn nhé :):p:p

[tex]PT \Leftrightarrow sinx+\frac{1}{2}(sin3x+sinx)+\sqrt[]{3}cos3x=2cos4x+2sin^3x[/tex]

\Leftrightarrow [tex]\frac{3}{2}sinx+\frac{1}{2}sin3x+\sqrt{3}cos3x=2cos4x+2sin^3x[/tex]

\Leftrightarrow[tex]3sinx+sin3x+2\sqrt[]{3}cos3x=4cos4x+4sin^3x[/tex]

\Leftrightarrow [tex]2sin3x+2\sqrt[]{3}cos3x=4cos4x[/tex]

\Leftrightarrow [tex]\frac{1}{2}sin3x+\frac{\sqrt[]{3}}{2}cos3x=cos4x[/tex]

\Leftrightarrow [tex]cos(\frac{pi}{6}-3x)=cos4x[/tex]

p/S: nhấn thanks nếu bạn thấy nó có ích nhé :D

:):):):)
 
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