Điều kiện xác định:

$\begin{array}{l} 1 + 2\sin 2x \ne 0\\ \Leftrightarrow \sin 2x \ne - \dfrac{1}{2} \Leftrightarrow \left[ \begin{array}{l} 2x \ne \dfrac{{7\pi }}{6} + k2\pi \\ 2x \ne - \dfrac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x \ne \dfrac{{7\pi }}{{12}} + k\pi \\ x \ne - \dfrac{\pi }{{12}} + k\pi \end{array} \right.\left( {k \in \mathbb Z} \right) \end{array}$

$\begin{array}{l} 5\left( {\sin x + \dfrac{{\cos 3x + \sin 3x}}{{1 + 2\sin 2x}}} \right) = 3 + \cos 2x\\  \Leftrightarrow 5\left( {\dfrac{{\sin x + 2\sin 2x\sin x + \cos 3x + \sin 3x}}{{1 + 2\sin 3x}}} \right) = 3 + \cos 2x\\  \Leftrightarrow 5\left( {\dfrac{{\sin x + \cos x - \cos 3x + \cos 3x + \sin 3x}}{{1 + 2\sin 3x}}} \right) = 3 + \cos 2x\\  \Leftrightarrow 5\left( {\dfrac{{\sin x + \cos x + \sin 3x}}{{1 + 2\sin 2x}}} \right) = 3 + \cos 2x\\  \Leftrightarrow 5\left( {\dfrac{{2\sin 2x\cos x + \cos x}}{{1 + 2\sin 2x}}} \right) = 3 + \cos 2x\\  \Leftrightarrow 5\left( {\dfrac{{\cos x\left( {2\sin 2x + 1} \right)}}{{1 + 2\sin 2x}}} \right) = 3 + \cos 2x\\  \Leftrightarrow 5\cos x = 3 + \cos 2x\\  \Leftrightarrow \cos 2x - 5\cos x + 3 = 0\\  \Leftrightarrow 2{\cos ^2}x - 5\cos x + 2 = 0\\  \Leftrightarrow \left[ \begin{array}{l} \cos x = 2(L)\\ \cos x = \dfrac{1}{2} \end{array} \right. \Leftrightarrow \cos x = \dfrac{1}{2}\\  \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{3} + k2\pi \\ x =  - \dfrac{\pi }{3} + k2\pi  \end{array} \right. \ \end{array}$

`        5(sinx +(cos3x+sin3x)/(1+2sin2x))=3+cos2x`(1)

Phương trình xác định `<=>1+2sin2x ne 0`

`<=>sin2x ne -1/2`

`<=>`\(\left[ \begin{array}{l}2x\ne-\frac{\pi}{6}+k2\pi(k \in\mathbb{Z})\\2x\ne \frac{7\pi}{6} +k2\pi(k \in \mathbb{Z})\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x\ne-\frac{\pi}{12}+k\pi(k \in\mathbb{Z})\\x\ne \frac{7\pi}{12} +k\pi(k \in \mathbb{Z})\end{array} \right.\) 

`=>`TXĐ: D=`RR text(\){-pi/12+kpi;(7pi)/12+kpi} (k in ZZ)`

(1)`<=>5(sinx+((4cos^3x-3cosx)+(3sinx-4sin^3x))/(1+2sin2x))=3+cos2x`

`<=>5(sinx+(4(cos^3x-sin^3x)+3(sinx-cosx))/(1+2sin2x))=3+cos2x`

`<=>5(sinx+(4(cosx-sinx)(sin^2x+sinxcosx+cos^2x)-3(cosx-sinx))/(1+2sin2x))=3+cos2x`

`<=>5(sinx+(4(cosx-sinx)(1+sinxcosx)-3(cosx-sinx))/(1+2sin2x))=3+cos2x`

`<=>5(sinx+((cosx-sinx)((4+4sinxcosx)-3))/(1+2sin2x))=3+cos2x`

`<=>5(sinx+((cosx-sinx)(1+2sin2x))/(1+2sin2x))=3+cos2x`

`<=>5(sinx+(cosx-sinx))=3+cos2x`

`<=>5cosx=3+(2cos^2x-1)`

`<=>2cos^2x-5cosx+2=0`

Đặt `t=cosx (t in [-1;1]text(\){(sqrt(6)+sqrt(2))/4;-(sqrt(6)+sqrt(2))/4})`

`=>2t^2-5t+2=0`

`<=>` \(\left[ \begin{array}{l}t=2\text(Loại)\\t=\frac{1}{2}\text(TM)\end{array} \right.\) 

`<=> cosx=1/2`

`<=>`\(\left[ \begin{array}{l}x=\frac{\pi}{3}+k2\pi(k \in \mathbb{Z})\\x=-\frac{\pi}{3}+k2\pi(k \in \mathbb{Z})\end{array} \right.\) 

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