ĐKXĐ: \(x<>\frac{\pi}{2}+k\pi\)

\(\sqrt3\cdot\cot^2x-4\cdot\cot x+\sqrt3=0\) (1)

\(\Delta=\left(-4\right)^2-2\cdot\sqrt3\cdot\sqrt3=16-2\cdot3=16-6=10>0\)

Do đó: (1) có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}\cot x=\frac{4-\sqrt{10}}{2\cdot\sqrt3}=\frac{4-\sqrt{10}}{\sqrt{12}}=\frac{4\sqrt3-\sqrt{30}}{6}\\ cotx=\frac{4+\sqrt{10}}{2\cdot\sqrt3}=\frac{4+\sqrt{10}}{\sqrt{12}}=\frac{4\sqrt3+\sqrt{30}}{6}\end{array}\right.\)

TH1: \(\cot x=\frac{4\sqrt3-\sqrt{30}}{6}\)

=>\(x=arc\cot\left(\frac{4\sqrt3-\sqrt{30}}{6}\right)+k\pi\) (nhận)

TH2: \(\cot x=\frac{4\sqrt3+\sqrt{30}}{6}\)

=>\(x=arc\cot\left(\frac{4\sqrt3+\sqrt{30}}{6}\right)+k\pi\) (nhận)

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

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

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

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

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

\(\Rightarrow x=...\)

2: \(cos\left(2x-\frac{\pi}{4}\right)+\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(cos\left(2x-\frac{\pi}{4}\right)=-\sin\left(x+\frac{\pi}{4}\right)=\sin\left(-x-\frac{\pi}{4}\right)\)

=>\(cos\left(2x-\frac{\pi}{4}\right)=cos\left(\frac{\pi}{2}+x+\frac{\pi}{4}\right)=cos\left(x+\frac34\pi\right)\)

=>\(\left[\begin{array}{l}2x-\frac{\pi}{4}=x+\frac34\pi+k2\pi\\ 2x-\frac{\pi}{4}=-x-\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x-x=\frac34\pi+\frac{\pi}{4}+k2\pi\\ 2x+x=-\frac34\pi+\frac{\pi}{4}+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\pi+k2\pi\\ 3x=-\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\pi+k2\pi\\ x=-\frac16\pi+\frac{k2\pi}{3}\end{array}\right.\)

3: \(\sin2x-\sin2x\cdot cosx=0\)

=>\(\sin2x\left(1-cosx\right)=0\)

TH1: sin 2x=0

=>\(2x=k\pi\)

=>\(x=\frac{k\pi}{2}\)

TH2: 1-cosx=0

=>cosx=1

=>\(x=k2\pi\)

1: \(\sin\left(2x-\frac{\pi}{4}\right)+\sin\left(3x+\frac{\pi}{3}\right)=0\)

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

=>\(\left[\begin{array}{l}3x+\frac{\pi}{3}=-2x+\frac{\pi}{4}+k2\pi\\ 3x+\frac{\pi}{3}=\pi+2x-\frac{\pi}{4}+k2\pi=\frac34\pi+2x+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}3x+2x=\frac{\pi}{4}-\frac{\pi}{3}+k2\pi=-\frac{\pi}{12}+k2\pi\\ 3x-2x=\frac34\pi-\frac{\pi}{3}+k2\pi=\frac{5}{12}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=-\frac{\pi}{12}+k2\pi\\ x=\frac{5}{12}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{\pi}{60}+\frac{k2\pi}{5}\\ x=\frac{5}{12}\pi+k2\pi\end{array}\right.\)

2: \(cos\left(2x-\frac{\pi}{4}\right)+\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(cos\left(2x-\frac{\pi}{4}\right)=-\sin\left(x+\frac{\pi}{4}\right)=\sin\left(-x-\frac{\pi}{4}\right)\)

=>\(cos\left(2x-\frac{\pi}{4}\right)=cos\left(\frac{\pi}{2}+x+\frac{\pi}{4}\right)=cosx\left(x+\frac34\pi\right)\)

=>\(\left[\begin{array}{l}2x-\frac{\pi}{4}=x+\frac34\pi+k2\pi\\ 2x-\frac{\pi}{4}=-x-\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac34\pi+\frac{\pi}{4}+k2\pi=\pi+k2\pi\\ 2x+x=-\frac34\pi+\frac{\pi}{4}+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\pi+k2\pi\\ 3x=-\frac{\pi}{2}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\pi+k2\pi\\ x=-\frac{\pi}{6}+\frac{k2\pi}{3}\end{array}\right.\)

1: \(\sin3x=-\frac12\)

=>\(\left[\begin{array}{l}3x=-\frac{\pi}{6}+k2\pi\\ 3x=\pi+\frac{\pi}{6}+k2\pi=\frac76\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac{\pi}{18}+\frac{k2\pi}{3}\\ x=\frac{7}{18}\pi+\frac{k2\pi}{3}\end{array}\right.\)

2: \(cos2x=-\frac{\sqrt2}{2}\)

=>\(\left[\begin{array}{l}2x=\frac34\pi+k2\pi\\ 2x=-\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac38\pi+k\pi\\ x=-\frac38\pi+k\pi\end{array}\right.\)

3: \(\sin\left(3x+45^0\right)=\frac{\sqrt3}{2}\)

=>\(\left[\begin{array}{l}3x+45^0=60^0+k\cdot360^0\\ 3x+45^0=180^0-60^0+k\cdot360^0=120^0+k\cdot360^0\end{array}\right.\)

=>\(\left[\begin{array}{l}3x=15^0+k\cdot360^0\\ 3x=75^0+k\cdot360^0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5^0+k\cdot120^0\\ x=25^0+k\cdot120^0\end{array}\right.\)

4: \(cos\left(2x+\frac{\pi}{6}\right)=\frac13\)

=>\(\left[\begin{array}{l}2x+\frac{\pi}{6}=arccos\left(\frac13\right)+k2\pi\\ 2x+\frac{\pi}{6}=-arccos\left(\frac13\right)+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=-\frac{\pi}{6}+arccos\left(\frac13\right)+k2\pi\\ 2x=-\frac{\pi}{6}-arccos\left(\frac13\right)+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{\pi}{12}+\frac12\cdot arccos\left(\frac13\right)+k\pi\\ x=-\frac{\pi}{12}-\frac12\cdot arccos\left(\frac13\right)+k\pi\end{array}\right.\)

1: \(\sin\left(2x-\frac{\pi}{4}\right)+\sin\left(3x+\frac{\pi}{3}\right)=0\)

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

=>\(\left[\begin{array}{l}3x+\frac{\pi}{3}=-2x+\frac{\pi}{4}+k2\pi\\ 3x+\frac{\pi}{3}=\pi+2x-\frac{\pi}{4}+k2\pi=\frac34\pi+2x+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}3x+2x=\frac{\pi}{4}-\frac{\pi}{3}+k2\pi=-\frac{\pi}{12}+k2\pi\\ 3x-2x=\frac34\pi-\frac{\pi}{3}+k2\pi=\frac{5}{12}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=-\frac{\pi}{12}+k2\pi\\ x=\frac{5}{12}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{\pi}{60}+\frac{k2\pi}{5}\\ x=\frac{5}{12}\pi+k2\pi\end{array}\right.\)

2: \(cos\left(2x-\frac{\pi}{4}\right)+\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(cos\left(2x-\frac{\pi}{4}\right)=-\sin\left(x+\frac{\pi}{4}\right)=\sin\left(-x-\frac{\pi}{4}\right)\)

=>\(cos\left(2x-\frac{\pi}{4}\right)=cos\left(\frac{\pi}{2}+x+\frac{\pi}{4}\right)=cosx\left(x+\frac34\pi\right)\)

=>\(\left[\begin{array}{l}2x-\frac{\pi}{4}=x+\frac34\pi+k2\pi\\ 2x-\frac{\pi}{4}=-x-\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac34\pi+\frac{\pi}{4}+k2\pi=\pi+k2\pi\\ 2x+x=-\frac34\pi+\frac{\pi}{4}+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\pi+k2\pi\\ 3x=-\frac{\pi}{2}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\pi+k2\pi\\ x=-\frac{\pi}{6}+\frac{k2\pi}{3}\end{array}\right.\)

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

=>\(\left[\begin{array}{l}2x-\frac{\pi}{5}=x+\frac{\pi}{5}+k2\pi\\ 2x-\frac{\pi}{5}=\pi-x-\frac{\pi}{5}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac25\pi+k2\pi\\ 3x=\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac25\pi+k2\pi\\ x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{array}\right.\)

4: \(\sin3x-cos2x=0\)

=>\(\sin3x=cos2x=\sin\left(\frac{\pi}{2}-2x\right)\)

=>\(\left[\begin{array}{l}3x=\frac{\pi}{2}-2x+k2\pi\\ 3x=\pi-\frac{\pi}{2}+2x+k2\pi=\frac{\pi}{2}+2x+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=\frac{\pi}{2}+k2\pi\\ x=\frac{\pi}{2}+k2\pi\end{array}\right.\)

=>\(x=\frac{\pi}{10}+\frac{k2\pi}{5}\)

5: \(\sin\left(\frac{x}{2}\right)\cdot cos\left(\frac{\pi}{3}\right)+\sin\left(\frac{\pi}{3}\right)\cdot cos\left(\frac{x}{2}\right)=\frac12\)

=>\(\sin\left(\frac{x}{2}+\frac{\pi}{3}\right)=\frac12\)

=>\(\left[\begin{array}{l}\frac{x}{2}+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\ \frac{x}{2}+\frac{\pi}{3}=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{x}{2}=\frac{\pi}{6}-\frac{\pi}{3}+k2\pi=-\frac{\pi}{6}+k2\pi\\ \frac{x}{2}=\frac56\pi-\frac{\pi}{3}+k2\pi=\frac{\pi}{2}+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{\pi}{3}+k4\pi\\ x=\pi+k4\pi\end{array}\right.\)

2: \(cos\left(2x-\frac{\pi}{4}\right)+\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(cos\left(2x-\frac{\pi}{4}\right)=-\sin\left(x+\frac{\pi}{4}\right)=\sin\left(-x-\frac{\pi}{4}\right)\)

=>\(cos\left(2x-\frac{\pi}{4}\right)=cos\left(\frac{\pi}{2}+x+\frac{\pi}{4}\right)=cos\left(x+\frac34\pi\right)\)

=>\(\left[\begin{array}{l}2x-\frac{\pi}{4}=x+\frac34\pi+k2\pi\\ 2x-\frac{\pi}{4}=-x-\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x-x=\frac34\pi+\frac{\pi}{4}+k2\pi\\ 2x+x=-\frac34\pi+\frac{\pi}{4}+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\pi+k2\pi\\ 3x=-\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\pi+k2\pi\\ x=-\frac16\pi+\frac{k2\pi}{3}\end{array}\right.\)

3: \(\sin2x-\sin2x\cdot cosx=0\)

=>\(\sin2x\left(1-cosx\right)=0\)

TH1: sin 2x=0

=>\(2x=k\pi\)

=>\(x=\frac{k\pi}{2}\)

TH2: 1-cosx=0

=>cosx=1

=>\(x=k2\pi\)

1: \(\sin\left(2x-\frac{\pi}{4}\right)+\sin\left(3x+\frac{\pi}{3}\right)=0\)

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

=>\(\left[\begin{array}{l}3x+\frac{\pi}{3}=-2x+\frac{\pi}{4}+k2\pi\\ 3x+\frac{\pi}{3}=\pi+2x-\frac{\pi}{4}+k2\pi=\frac34\pi+2x+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}3x+2x=\frac{\pi}{4}-\frac{\pi}{3}+k2\pi=-\frac{\pi}{12}+k2\pi\\ 3x-2x=\frac34\pi-\frac{\pi}{3}+k2\pi=\frac{5}{12}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=-\frac{\pi}{12}+k2\pi\\ x=\frac{5}{12}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{\pi}{60}+\frac{k2\pi}{5}\\ x=\frac{5}{12}\pi+k2\pi\end{array}\right.\)