\(\displaystyle{\sin{{2}}}{x}={\sin{{x}}}\)

\(\displaystyle{2}{\sin{{x}}}{\cos{{x}}}={\sin{{x}}}\)

\(\displaystyle{2}{\sin{{x}}}{\cos{{x}}}-{\sin{{x}}}={0}\)

\(\displaystyle{\sin{{x}}}{\left({2}{\cos{{x}}}-{1}\right)}={0}\)

Solution A: \(\displaystyle{\sin{{x}}}={0}\Rightarrow{x}={k}\pi,{k}\in{\mathbb{{{Z}}}}\)

Solution B: \(\displaystyle{2}{\cos{{x}}}={1}\Rightarrow{\cos{{x}}}={\frac{{12}}{,}}{x}=\pm{\frac{{\pi}}{{{3}}}}+{2}{k}\pi={\frac{{\pi}}{{{3}}}}+{2}{k}\pi={\frac{{\pi}}{{{3}}}}{\left({6}{k}\pm{1}\right)},\ {k}\in{\mathbb{{{Z}}}}\)

\(\displaystyle\therefore{x}={k}\pi\) or \(\displaystyle{x}={\frac{{\pi}}{{{3}}}}{\left({6}{k}\pm{1}\right)},{k}\in{\mathbb{{{Z}}}}\)

\(\displaystyle{2}{\sin{{x}}}{\cos{{x}}}={\sin{{x}}}\)

\(\displaystyle{2}{\sin{{x}}}{\cos{{x}}}-{\sin{{x}}}={0}\)

\(\displaystyle{\sin{{x}}}{\left({2}{\cos{{x}}}-{1}\right)}={0}\)

Solution A: \(\displaystyle{\sin{{x}}}={0}\Rightarrow{x}={k}\pi,{k}\in{\mathbb{{{Z}}}}\)

Solution B: \(\displaystyle{2}{\cos{{x}}}={1}\Rightarrow{\cos{{x}}}={\frac{{12}}{,}}{x}=\pm{\frac{{\pi}}{{{3}}}}+{2}{k}\pi={\frac{{\pi}}{{{3}}}}+{2}{k}\pi={\frac{{\pi}}{{{3}}}}{\left({6}{k}\pm{1}\right)},\ {k}\in{\mathbb{{{Z}}}}\)

\(\displaystyle\therefore{x}={k}\pi\) or \(\displaystyle{x}={\frac{{\pi}}{{{3}}}}{\left({6}{k}\pm{1}\right)},{k}\in{\mathbb{{{Z}}}}\)