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

\(\Leftrightarrow\left[{}\begin{matrix}4x=-\dfrac{\pi}{2}+k2\pi\\6x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+k\dfrac{\pi}{2}\\x=\dfrac{7\pi}{36}+k\dfrac{\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)

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

\(\Leftrightarrow4sinx+\sqrt{3}cosx-2\sqrt{3}sinx.cosx-2=0\)

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

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

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

\(\Leftrightarrow...\)

a) \(sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(x-\dfrac{\pi}{3}\right)=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+k\pi\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{7\pi}{6}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{6}+\left(k+1\right)\pi\end{matrix}\right.\)

c: =>\(cos\left(x-\dfrac{pi}{6}\right)=-sin\left(2x+\dfrac{pi}{3}\right)\)

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

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

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

=>\(\left[{}\begin{matrix}-2x-\dfrac{pi}{3}=-x+\dfrac{2}{3}pi+k2pi\\-2x-\dfrac{pi}{3}=pi+x-\dfrac{2}{3}pi+k2pi\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}-x=pi+k2pi\\-3x=\dfrac{2}{3}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-pi-k2pi\\x=-\dfrac{2}{9}pi-\dfrac{k2pi}{3}\end{matrix}\right.\)

Lời giải:ĐK: $\cos 3x>\frac{-1}{2}$

PT $\Rightarrow 4\sin ^2\frac{x}{2}-\sqrt{3}\cos 2x-1-2\cos ^2(x-\frac{3\pi}{4})=0$

$\Leftrightarrow 2(1-\cos x)-\sqrt{3}\cos 2x-2+[1-2\cos ^2(x-\frac{3\pi}{4})]=0$

$\Leftrightarrow -2\cos x-\sqrt{3}\cos 2x-cos (2x-\frac{3\pi}{2})=0$

$\Leftrightarrow 2\cos x+\sqrt{3}\cos 2x+\cos (2x-\frac{3\pi}{2})=0$

$\Leftrightarrow 2\cos x+\sqrt{3}\cos 2x+\sin 2x=0$

$\Leftrightarrow \cos x+\frac{\sqrt{3}}{2}\cos 2x+\frac{1}{2}\sin 2x=0$

$\Leftrightarrow \cos x-\cos (2x+\frac{5\pi}{6})=0

$\Leftrightarrow \cos x=\cos (2x+\frac{5\pi}{6})$

$\Rightarrow x+2k\pi =2x+\frac{5}{6}\pi$ hoặc $-x+2k\pi =2x+\frac{5}{6}\pi$

Vậy......

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

\(\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\left(k\in Z\right)\)

`sin (x-\pi/3)=1`

`<=>x-\pi/3=\pi/2+k2\pi` , `k in ZZ`

`<=>x=[5\pi]/6 +k2\pi` , `k in ZZ`

Vậy ptr có nghiệm `x=[5\pi]/6 +k2\pi` , `k in ZZ`

\(\sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\pi}{2}\Leftrightarrow x-\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{3\pi}{4}+k2\pi\left(k\in Z\right)\)

`sin(x- (pi)/4) = (pi)/2`

`<=> x - (pi)/4 = (pi)/2 + k2(pi)`

`<=> x = (3(pi))/4 + k2(pi)`.

a.

\(sin\left(2x+1\right)=-cos\left(3x-1\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+2+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

b.

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)

c.

\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=-sin\left(x-\frac{2\pi}{5}-\pi\right)\)

\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=sin\left(x-\frac{2\pi}{5}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{2\pi}{3}=x-\frac{2\pi}{5}+k2\pi\\3x+\frac{2\pi}{3}=\frac{7\pi}{5}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{8\pi}{15}+k\pi\\x=\frac{11\pi}{60}+\frac{k\pi}{2}\end{matrix}\right.\)

d.

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

\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)

\(\sin\left(x-\dfrac{\pi}{2}\right)=1\Leftrightarrow x-\dfrac{\pi}{2}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\pi+k2\pi\left(k\in Z\right)\)

\(\sin\left(2x+\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{4}=-\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{4}=\pi-\left(-\dfrac{\pi}{6}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{6}-\dfrac{\pi}{4}+k2\pi\\2x=\pi+\dfrac{\pi}{6}-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{5\pi}{12}+k2\pi\\2x=\dfrac{11\pi}{12}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5\pi}{24}+k\pi\\x=\dfrac{11\pi}{24}+k\pi\end{matrix}\right.\left(k\in Z\right)\)

=)) sẽ chỉ có 1 người trl:vv