Lượng giác khó

T

trantien.hocmai

$16sinxcos4x+1=\sqrt{3}(tanx+tan2x)+tanx.tan2x$
$ \leftrightarrow 16sinxcos4x+1=\sqrt{3}.\frac{sinxcos2x+sin2xcosx}{cosxcos2x}+\frac{sinxsin2x}{cosxcos2x}$
$ \leftrightarrow 16sinx.cosx.cos2x.cos4x+cosx.cos2x=\sqrt{3}sin3x+sinx.sin2x$
$ \leftrightarrow 8sin2x.cos2x.cos4x-\sqrt{3}sin2x=sinx.sin2x-cosx.cos2x$
$ \leftrightarrow 4sin4x.cos4x-\sqrt{3}sỉnx+cosx.cos2x-sinx.sin2x=0$
$ \leftrightarrow 2sin8x-\sqrt{3}sin3x+cos3x=0$
đến đây thì dễ rồi nhá
 
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