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

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

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

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

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

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

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/

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

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

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

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

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

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

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

a, \(sin4x.cosx-sin3x=0\)

\(\Leftrightarrow\dfrac{1}{2}sin5x+\dfrac{1}{2}sin3x-sin3x=0\)

\(\Leftrightarrow sin5x=sin3x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\end{matrix}\right.\)

b, \(sin2x+\sqrt{3}cos2x=\sqrt{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=\dfrac{\pi}{4}+k2\pi\\2x+\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{24}+k\pi\\x=\dfrac{5\pi}{24}+k\pi\end{matrix}\right.\)

12545789

`cos 2x+\sqrt{3}sin 2x+\sqrt{3}sin x-cos x=4`

`<=>1/2 cos 2x+\sqrt{3}/2 sin 2x+\sqrt{3}/2 sin x-1/2 cos x=2`

`<=>sin(\pi/6 +2x)+sin(x-\pi/6)=2`

Vì `-1 <= sin (\pi/6 +2x) <= 1`

     `-1 <= sin (x-\pi/6) <= 1`

 Dấu "`=`" xảy ra `<=>{(sin(\pi/6+2x)=1),(sin(x-\pi/6)=1):}`

        `<=>{(\pi/6+2x=\pi/2+k2\pi),(x-\pi/6=\pi/2+k2\pi):}`

        `<=>{(x=\pi/6+k\pi),(x=[2\pi]/3+k2\pi):}`    `(k in ZZ)`

Từ phương trình ban đầu ta có : \(2\cos5x\sin x=\sqrt{3}\sin^2x+\sin x\cos x\)

                                                \(\Leftrightarrow\begin{cases}\sin x=0\\2\cos5x=\sqrt{3}\sin x+\cos x\end{cases}\)

+) \(\sin x=0\Leftrightarrow x=k\pi\)

+)\(2\cos5x=\sqrt{3}\sin x+\cos x\Leftrightarrow\cos5x=\cos\left(x-\frac{\pi}{3}\right)\)

                                             \(\Leftrightarrow\begin{cases}x=-\frac{\pi}{12}+\frac{k\pi}{2}\\x=\frac{\pi}{18}+\frac{k\pi}{3}\end{cases}\)

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