$tan(2x-\frac{\pi}{4})tan(2x+\frac{\pi}{4})=\frac{4(cos2x)^{2}}{tanx-cotx}$
Giải phương trình $tan(2x-\frac{\pi}{4})tan(2x+\frac{\pi}{4})=\frac{4(cos2x)^{2}}{tanx-cotx}$
Started By KieuOanh1996, 09-06-2014 - 22:01
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Posted 13-06-2014 - 16:00
duaconcuachua98
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$tan(2x-\frac{\pi}{4})tan(2x+\frac{\pi}{4})=\frac{4(cos2x)^{2}}{tanx-cotx}$
Phương trình tương đương $\frac{tan^{2}2x-tan^{2}\frac{\pi }{4}}{1-tan^{2}2x.tan^{2}\frac{\pi }{4}}=\frac{2cos^{2}2x}{cot2x}\Leftrightarrow 2sin2xcos2x=-1\Leftrightarrow sin4x=-1\Leftrightarrow 4x=\frac{-\pi }{2}+k2\pi \Leftrightarrow x=\frac{-\pi }{8}+\frac{k\pi }{2}$
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