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lovelycat_handoi95

[TEX]\frac{\sqrt{3}-4sin(2x+\frac{\pi }{3})+2sin4x}{sin(x-\frac{\pi }{3})}=6sin^2x-2cos^2x.[/TEX]

[TEX]PT \Leftrightarrow \frac{\sqrt{3}-2sin2x-2\sqrt{3}cos2x+2sin4x}{{\frac{1}{2}}sin2x-{\frac{\sqrt{3}}{2}}cos2x} = 3(1-cos2x)-(cos2x+1) \\ \Leftrightarrow \frac{2sin2x(2cos2x-1)-\sqrt{3}(2cos2x-1) }{{\frac{1}{2}}sin2x-{\frac{\sqrt{3}}{2}}cos2x}= -2(2cos2x-1) \\ \Leftrightarrow (2cos2x-1)(2sin2x-\sqrt{3})=(2cos2x-1)(\sqrt{3}cos2x-sin2x) \\ \Leftrightarrow (2cos2x-1)(3sin2x-\sqrt{3}-\sqrt{3}cos2x)=0 [/TEX]
 
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