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Question

$(tanxcot2x-1)sin(4x+\frac{\Pi }{2})=\frac{-1}{2}(\sin^{4} +\cos^{4} x)$

gunbk92 · Answer 1 · 3/18/2013 9:56:20 AM

Đk:

$\sin 2x \neq 0$

$\Leftrightarrow (\frac{\sin x\cos 2x-\cos x\sin 2x}{\cos x\sin 2x})\cos 4x=\frac{-1}{2}(\sin^4x+\cos^4x)$

$\Leftrightarrow \frac{-\sin x}{2\cos^2x\sin x}\cos4x=\frac{-1}{2}[(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x]$

$\Leftrightarrow \cos 4x=\cos^2x(1-\frac{1}{2}\sin^22x)$

$\Leftrightarrow 2\cos^22x-1=(\frac{\cos2x+1}{2})(\frac{1}{2}+\frac{1}{2}\cos^22x)$ (1)

Đặt

$t = \cos 2x;\sin 2x\neq0\Rightarrow \cos 2x\neq\pm 1\Rightarrow -1<t<1$

$(1)\Leftrightarrow 2t^2-1=(\frac{t+1}{2})\frac{1}{2}(1+t^2)$

$\Leftrightarrow t^3-7t^2+t+5=0$

$\Leftrightarrow \left[ \begin{matrix} \begin{matrix} t=1(loại)\\ t=3-\sqrt{14} \end{matrix}\\ t=3+\sqrt{14}(loại)\end{matrix} \right.$

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