a: \(2\cdot cos\left(x-\frac{\pi}{3}\right)-1=0\)

=>\(2\cdot cos\left(x-\frac{\pi}{3}\right)=1\)

=>\(cos\left(x-\frac{\pi}{3}\right)=\frac12\)

=>\(\left[\begin{array}{l}x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\ x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{3}+\frac{\pi}{3}+k2\pi=\frac23\pi+k2\pi\\ x=k2\pi\end{array}\right.\)

b: \(\sin2x=\frac{\sqrt3}{2}\)

=>\(\left[\begin{array}{l}2x=\frac{\pi}{3}+k2\pi\\ 2x=\pi-\frac{\pi}{3}+k2\pi=\frac23\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+k\pi\\ x=\frac{\pi}{3}+k\pi\end{array}\right.\)

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

=>\(cos\left(\frac12x+\frac{\pi}{6}\right)=cos\left(\frac{\pi}{2}-x\right)\)

=>\(\left[\begin{array}{l}\frac12x+\frac{\pi}{6}=\frac{\pi}{2}-x+k2\pi\\ \frac12x+\frac{\pi}{6}=x-\frac{\pi}{2}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac12x+x=\frac{\pi}{2}-\frac{\pi}{6}+k2\pi\\ \frac12x-x=-\frac{\pi}{2}-\frac{\pi}{6}+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac32x=\frac{\pi}{3}+k2\pi\\ -\frac12x=-\frac43\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac29\pi+\frac{k4\pi}{3}\\ x=\frac83\pi-k4\pi\end{array}\right.\)

d: \(\cot\left(x+\frac{\pi}{6}\right)=7\)

=>\(x+\frac{\pi}{6}=arc\cot\left(7\right)+k\pi\)

=>\(x=-\frac{\pi}{6}+arc\cot\left(7\right)+k\pi\)

e: \(\tan\left(2x-\frac34\pi\right)=1\)

=>\(2x-\frac34\pi=\frac{\pi}{4}+k\pi\)

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

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

g: \(\sin\left(x+\frac{\pi}{4}\right)=1\)

=>\(x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac{\pi}{2}-\frac{\pi}{4}+k2\pi=\frac{\pi}{4}+k2\pi\)

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

=>\(\frac12x+\frac{\pi}{4}=\frac{\pi}{2}+k\pi\)

=>\(\frac12x=\frac14\pi+k\pi\)

=>\(x=\frac12\pi+k2\pi\)

i: \(\sqrt2\cdot\sin\left(x+\frac23\pi\right)=-1\)

=>\(\sin\left(x+\frac23\pi\right)=-\frac{1}{\sqrt2}\)

=>\(\left[\begin{array}{l}x+\frac23\pi=-\frac{\pi}{4}+k2\pi\\ x+\frac23\pi=\pi+\frac{\pi}{4}+k2\pi=\frac54\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac{\pi}{4}-\frac23\pi+k2\pi=-\frac{11}{12}\pi+k2\pi\\ x=\frac54\pi-\frac23\pi+k2\pi=\frac{7}{12}\pi+k2\pi\end{array}\right.\)

k: \(4\cdot cos\left(\frac{x}{2}+\frac{\pi}{4}\right)=5\)

=>\(cos\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac54>1\)

=>x∈∅

l: \(\sqrt2\cdot cos\left(x+\frac{\pi}{3}\right)+1=0\)

=>\(\sqrt2\cdot cos\left(x+\frac{\pi}{3}\right)=-1\)

=>\(cos\left(x+\frac{\pi}{3}\right)=-\frac12\)

=>\(\left[\begin{array}{l}x+\frac{\pi}{3}=\frac23\pi+k2\pi\\ x+\frac{\pi}{3}=-\frac23\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\pi+k2\pi\\ x=-\pi+k2\pi\end{array}\right.\)

m: \(cos\left(x+\frac{\pi}{3}\right)=\frac{1}{\sqrt2}\)

=>\(\left[\begin{array}{l}x+\frac{\pi}{3}=\frac{\pi}{4}+k2\pi\\ x+\frac{\pi}{3}=-\frac{\pi}{4}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{4}-\frac{\pi}{3}+k2\pi=-\frac{\pi}{12}+k2\pi\\ x=-\frac{\pi}{4}-\frac{\pi}{3}+k2\pi=-\frac{7}{12}\pi+k2\pi\end{array}\right.\)