Giải các pt sau :
1) $\frac{2\sqrt{3}cos^{3}2x+2sin^{3}2x}{\sqrt{3}cos2x-sin2x}=cos4x$
2)$\frac{40(sin^{3\frac{x}{2}}-cos^{3}\frac{x}{2})}{16sin\frac{x}{2}-25sin\frac{x}{2}}=sinx$$\Leftrightarrow \sqrt{3}cos^{3}2x+sin^{3}2x+sin2xcos^{2}2x+\sqrt{3}sin^{2}2xcos2x=0$
3) $\frac{1}{sinx}-\frac{1}{sin2x}= \frac{1}{sin4x}$
4) $\frac{1-sin^{6}-cos^{6}x}{1-sin^{4}x-cos^{4}x}=2cos^{2}3x$
1) Điều kiện...........
$\Leftrightarrow 2\sqrt{3}cos^{3}2x+2sin^{3}2x=cos4x(\sqrt{3}cos2x-sin2x)$
$\Leftrightarrow 2\sqrt{3}cos^{3}x+2sin^{3}2x=(cos^{2}2x-sin^{2}2x)(\sqrt{3}cos2x-sin2x)$
$\Leftrightarrow 2\sqrt{3}cos^{3}2x+2sin^{3}2x=\sqrt{3}cos^{3}2x-sin2xcos^{2}2x-\sqrt{3}sin^{2}2xcos2x-sin^{3}2x$
$\Leftrightarrow \sqrt{3}cos^{3}2x+cos^{2}2xsin2x+\sqrt{3}sin^{2}2xcos2x+sin^{3}2x=0$
$\Leftrightarrow \sqrt{3}cos2x(cos^{2}2x+sin^{2}2x)+sin2x(cos^{2}2x+sin^{2}2x)=0$
$\Leftrightarrow \sqrt{3}cos2x+sin2x=0\Rightarrow tan2x=-\sqrt{3}$
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