\(\sqrt{2}\left(2cos^2x-3sin2x\right)=4cosx.sin2x+2\left(sinx-cosx\right)\)

\(\Leftrightarrow\left(2\sqrt{2}cos^2x+2cosx\right)-3\sqrt{2}sin2x-4cosx.sin2x-2sinx=0\)

\(\Leftrightarrow2cosx\left(\sqrt{2}cosx+1\right)-6\sqrt{2}sinx.cosx-4cosx^2.sinx-2sinx=0\)

\(\Leftrightarrow2cosx\left(\sqrt{2}cosx+1\right)-2sinx\left(4cos^2x+3\sqrt{2}cosx+1\right)=0\)

\(\Leftrightarrow2cosx\left(\sqrt{2}cosx+1\right)-2sinx\left(\sqrt{2}cosx+1\right)\left(2\sqrt{2}cosx+1\right)=0\)

\(\Leftrightarrow\left(2cosx-4\sqrt{2}cosx.sinx-2sinx\right)\left(\sqrt{2}cosx+1\right)=0\)

\(\Leftrightarrow\left[2\sqrt{2}-2\sqrt{2}\left(cosx-sinx\right)^2+2\left(cosx-sinx\right)\right]\left(\sqrt{2}cosx+1\right)=0\)

Đặt \(t=cosx-sinx\left(t\in\left[-\sqrt{2};\sqrt{2}\right]\right)\)

\(pt\Leftrightarrow\left[{}\begin{matrix}cosx=-\dfrac{1}{\sqrt{2}}\\\sqrt{2}t^2-t-\sqrt{2}=0\end{matrix}\right.\)

...

1.

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

Pt trở thành:

\(t^3+\frac{t^2-1}{2}-1=0\)

\(\Leftrightarrow2t^3+t^2-3=0\)

\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)

\(\Leftrightarrow t=1\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(t^4-3\left(t^2-1\right)-1=0\)

\(\Leftrightarrow t^4-3t^2+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t^2=1\\t^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}1+sin2x=1\\1+sin2x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow...\)

3.

\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)+2\left(sinx+cosx\right)-6sinx.cosx=0\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

Pt trở thành:

\(t\left(1-\frac{t^2-1}{2}\right)+2t-3\left(t^2-1\right)=0\)

\(\Leftrightarrow-t^3-6t^2+7t+6=0\)

Nghiệm của pt bậc 3 này rất xấu, chắc bạn ghi ko đúng đề bài

ĐKXĐ: ...

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx=\dfrac{3}{2}\left(1+tan^2x\right)-\sqrt{3}tanx\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=\dfrac{3}{2}\left(tanx-\dfrac{\sqrt{3}}{3}\right)^2+1\)

\(\left\{{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)\le1\\\dfrac{3}{2}\left(tanx-\dfrac{\sqrt{3}}{3}\right)^2+1\ge1\end{matrix}\right.\)

Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)=1\\tanx=\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{\pi}{6}+k2\pi\)

\(\Leftrightarrow2\left(\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx\right)+2cos\left(x-\dfrac{\pi}{3}\right)=2\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)+cos\left(x-\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{3}+k2\pi\\x=k2\pi\end{matrix}\right.\)

Ta có : \(2cos^2x+2\sqrt{3}sinx.cosx+1=3\left(sinx+\sqrt{3}cosx\right)\) 

\(\Leftrightarrow3cos^2x+sin^2x+2\sqrt{3}sinxcosx=3\left(sinx+\sqrt{3}cosx\right)\) 

\(\Leftrightarrow\left(\sqrt{3}cosx+sinx\right)^2=3\left(\sqrt{3}cosx+sinx\right)\) 

\(\Leftrightarrow\left(\sqrt{3}cosx+sinx\right)\left(\sqrt{3}cosx+sinx-3\right)=0\) 

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3}cosx+sinx=0\\\sqrt{3}cos+sinx=3\end{matrix}\right.\) 

Thấy : \(-1\le sinx;cosx\le1\Rightarrow\sqrt{3}cosx+sinx\le1+\sqrt{3}< 3\) 

Do đó : \(\sqrt{3}cosx+sinx=0\)  \(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx=0\)

\(\Leftrightarrow sin\dfrac{\pi}{3}.cosx+cos\dfrac{\pi}{3}sinx=0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\) ( k thuộc Z ) 

Vậy ... 

\(\Leftrightarrow cos3x+\sqrt{3}sin3x=\sqrt{3}cosx+sinx\)

\(\Leftrightarrow\dfrac{1}{2}cos3x+\dfrac{\sqrt{3}}{2}sin3x=\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx\)

\(\Leftrightarrow cos\left(3x-\dfrac{\pi}{3}\right)=cos\left(x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\\3x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+k2\pi\\x\ne-\dfrac{\pi}{6}+k2\pi\\x\ne\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\dfrac{cosx-2sinx.cosx}{1-2sin^2x+sinx}=\sqrt{3}\)

\(\Leftrightarrow\dfrac{cosx-sin2x}{cos2x+sinx}=\sqrt{3}\)

\(\Rightarrow cosx-sin2x=\sqrt{3}cos2x+\sqrt{3}sinx\)

\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)

\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos\left(2x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=x+\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{6}=-x-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\left(loại\right)\\x=-\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

ĐKXĐ : \(sinx\ne1;-\dfrac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+2k\pi\\x\ne\dfrac{-\pi}{6}+2k\pi;\dfrac{7\pi}{6}+2k\pi\end{matrix}\right.\)   

\(\Leftrightarrow x\ne\dfrac{-\pi}{6}+\dfrac{2}{3}k\pi\)( k thuộc Z ) 

P/t đã cho \(\Leftrightarrow\dfrac{cosx-sin2x}{1-2sin^2x+sinx}=\sqrt{3}\) 

\(\Leftrightarrow cosx-sin2x=\sqrt{3}\left(cos2x+sinx\right)\)

\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)

\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos\left(2x+\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+\dfrac{\pi}{3}+2k\pi\\2x+\dfrac{\pi}{6}=-x-\dfrac{\pi}{3}+2k\pi\end{matrix}\right.\) ( k thuộc Z ) 

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+2k\pi\\x=\dfrac{-\pi}{6}+\dfrac{2}{3}k\pi\left(L\right)\end{matrix}\right.\)

Vậy ...