[toán 11]Giải pt lượng giác

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truongduong9083

$$\tan^2x+\cot^2x+\cot^2{2x} = \dfrac{11}{3}$$$$ \Longleftrightarrow \dfrac{\sin^2x}{\cos^2x}+\dfrac{\cos^2x}{\sin^2x}+\dfrac{\cos^22x}{\sin^22x}=\dfrac{11}{3}$$$$ \Longleftrightarrow 4\sin^4x+4\cos^4x+\cos^22x=\dfrac{11}{3}\sin^22x$$$$ \Longleftrightarrow 4-8\sin^2x\cos^2x+\cos^22x=\dfrac{11}{3}\sin^22x$$$$ \Longleftrightarrow 4-2\sin^22x+1-\sin^22x=\dfrac{11}{3}\sin^22x$$$$ \Longleftrightarrow \sin^22x=\dfrac{3}{4}$$$$ \Longleftrightarrow \cos4x=\dfrac{-1}{2}$$
 
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