Thấy cosx= 0 là nghiệm của phương trình => \(x=\dfrac{\pi}{2}+k\pi\)

Xét cosx khác 0, chia cả 2 vế cho cos^2 x

\(\Leftrightarrow\tan^2x-\sqrt{3}\tan x+2=1+\tan^2x\)

\(\Leftrightarrow\tan x=\dfrac{\sqrt{3}}{3}\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

\(\sqrt{3}sin2x-cos2x=\sqrt{2}\)

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

\(\Leftrightarrow cos\left(\dfrac{\pi}{6}\right)sin2x-sin\left(\dfrac{\pi}{6}\right)cos2x=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{4}\right)\)

Làm nốt

\(\sqrt{2}sinx+sin2x=\sqrt{3}cos2x-\sqrt{6}cosx\)

\(\Leftrightarrow\dfrac{\sqrt{2}}{2}sinx+\dfrac{1}{2}sin2x-\dfrac{\sqrt{3}}{2}cos2x+\dfrac{\sqrt{6}}{2}cosx=0\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{3}\right)+sin\left(2x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{6}\right)+2sin\left(x-\dfrac{\pi}{6}\right).cos\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{6}\right)\left[1+\sqrt{2}sin\left(x-\dfrac{\pi}{6}\right)\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{6}\right)=0\\sin\left(x-\dfrac{\pi}{6}\right)=-\dfrac{1}{\sqrt{2}}\end{matrix}\right.\)

Đến đấy thì dễ rồi.

\(\Leftrightarrow\sqrt{2}\left(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\right)+\dfrac{1}{2}sin2x-\dfrac{\sqrt{3}}{2}cos2x=0\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{3}\right)+sin\left(2x-\dfrac{\pi}{3}\right)=0\)

Đặt \(x+\dfrac{\pi}{3}=u\Rightarrow2x-\dfrac{\pi}{3}=2u-\pi\)

\(\Rightarrow\sqrt{2}sinu+sin\left(2u-\pi\right)=0\)

\(\Leftrightarrow\sqrt{2}sinu-sin2u=0\)

\(\Leftrightarrow sinu\left(\sqrt{2}-2cosu\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=0\\cosu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)=0\\cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

a/

\(\Leftrightarrow2sinx.cosx-2\sqrt{3}cos^2x-4cosx=0\)

\(\Leftrightarrow2cosx\left(sinx-\sqrt{3}cosx-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\sinx-\sqrt{3}cosx=2\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=1\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=1\)

\(\Leftrightarrow x-\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{5\pi}{6}+k2\pi\)

b/

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

\(\Leftrightarrow\left(sinx-\sqrt{3}cosx\right)\left(sinx+\sqrt{3}cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\sqrt{3}cosx\left(1\right)\\sinx+\sqrt{3}cosx=1\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow tanx=\sqrt{3}\)

\(\Rightarrow x=\frac{\pi}{3}+k\pi\)

\(\left(2\right)\Leftrightarrow\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=\frac{1}{2}\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

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

c/

\(\Leftrightarrow sin6x\left(cos3x-1-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin6x=0\Rightarrow x=\frac{k\pi}{6}\\cos3x-sin3x=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow sin3x-cos3x=-1\)

\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)=-1\)

\(\Leftrightarrow sin\left(3x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\3x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

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

ĐKXĐ: \(sinx\ne\dfrac{\sqrt{2}}{2}\)

\(\left(sinx-cosx\right)\left(sin2x-3\right)+\left(sinx-cosx\right)^2+\left(sin^2x-cos^2x\right)=0\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(sin2x-3\right)+\left(sinx-cosx\right)^2+\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(sin2x-3+2sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\\left(sin2x-1\right)+2\left(sinx+1\right)=0\left(vô-nghiệm\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)

Kết hợp ĐKXĐ \(\Rightarrow x=-\dfrac{\pi}{4}+k2\pi\)

\(\Leftrightarrow-cos2x+\sqrt{3}sin2x=\sqrt{3}-1\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=\frac{\sqrt{3}-1}{2}\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=\frac{\sqrt{3}-1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=arcsin\left(\frac{\sqrt{3}-1}{2}\right)+k2\pi\\2x-\frac{\pi}{6}=\pi-arcsin\left(\frac{\sqrt{3}-1}{2}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{1}{2}arcsin\left(\frac{\sqrt{3}-1}{6}\right)+k\pi\\x=\frac{7\pi}{12}-\frac{1}{2}arcsin\left(\frac{\sqrt{3}-1}{6}\right)+k\pi\end{matrix}\right.\)

Đặt tanx = t

sin2x = 2sinx.cosx

= 2. tanx . cos²x

= \(2tanx.\dfrac{1}{tan^2x+1}\)

= \(\dfrac{2t}{t^2+1}\)

Vậy ta có phương trình

\(\dfrac{2t}{t^2+1}+4t=\dfrac{9\sqrt{3}}{2}\)