Giải pt sau :

X

xuanquynh97

ĐK $sin2x \not=0$

PT \Leftrightarrow $\dfrac{\sqrt{3}}{cos^2x}-2\sqrt{3}=2cotx+2-\dfrac{4}{sin2x}-2$


\Leftrightarrow $\sqrt{3}(\dfrac{1}{cos^2x}-2)=2(\dfrac{cosx}{sinx}-\dfrac{1}{sinxcosx})$

\Leftrightarrow $\sqrt{3}.\dfrac{cos2x}{cos^2x}=2.\dfrac{sinx}{cosx}$

\Leftrightarrow $\sqrt{3}.cos2x-sin2x=0$
 
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