$\sin x^{4} +cosx^{4} \sin2x =1/2.(\tan x+\cot x)$
Đề có phải là $\dfrac{\sin^4 x+\cos^4 x}{\sin 2x} =\dfrac{1}{2}.(\tan x +\cot x)$
Điều kiện $\sin 2x \ne 0 \Leftrightarrow x\ne \dfrac{k\pi}{2};\ k\in Z$

PT$ \Leftrightarrow \dfrac{\sin^4 x+\cos^4 x}{\sin 2x}=\dfrac{1}{2}. (\dfrac{\sin^2 x +\cos^2 x}{\sin x \cos x})$

$\Leftrightarrow \sin^4 x +\cos^4 x = 1$
$\Leftrightarrow 1-2sin^{2}x.cos^{2}x=1$
$\Leftrightarrow sin^{2}2x=0$
$\Leftrightarrow sin2x=0$
Điều kiện $\sin 2x \ne 0 \Leftrightarrow x\ne \dfrac{k\pi}{2};\ k\in Z$

PT$ \Leftrightarrow \dfrac{\sin^4 x+\cos^4 x}{\sin 2x}=\dfrac{1}{2}. (\dfrac{\sin^2 x +\cos^2 x}{\sin x \cos x})$

$\Leftrightarrow \sin^4 x +\cos^4 x = 1$

$\Leftrightarrow (1-\cos^2 x)^2 = (1-\cos^2 x)(1+\cos^2 x)$

$\Leftrightarrow (1-\cos^2 x) .[1-\cos^2 x-1-\cos^2 x]=0$ dễ rồi tự làm đi
Điều kiện sin2x0xkπ2; kZ

PTsin4x+cos4xsin2x=12.(sin2x+cos2xsinxcosx)

sin4x+cos4x=1 


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