# How do you solve \sin2x=\sin x?

How do you solve $$\displaystyle{\sin{{2}}}{x}={\sin{{x}}}$$?

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Philip Williams
$$\displaystyle{\sin{{2}}}{x}={\sin{{x}}}$$
$$\displaystyle{2}{\sin{{x}}}{\cos{{x}}}={\sin{{x}}}$$
$$\displaystyle{2}{\sin{{x}}}{\cos{{x}}}-{\sin{{x}}}={0}$$
$$\displaystyle{\sin{{x}}}{\left({2}{\cos{{x}}}-{1}\right)}={0}$$
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 $$\displaystyle{x}={\frac{{\pi}}{{{3}}}}{\left({6}{k}\pm{1}\right)},{k}\in{\mathbb{{{Z}}}}$$
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Paineow
Factor $$\displaystyle{\sin{{\left({x}\right)}}}$$ out of $$\displaystyle{2}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}-{\sin{{\left({x}\right)}}}$$
$$\displaystyle{\sin{{\left({x}\right)}}}{\left({2}{\cos{{\left({x}\right)}}}-{1}\right)}={0}$$
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
$$\displaystyle{\sin{{\left({x}\right)}}}={0}$$
$$\displaystyle{2}{\cos{{\left({x}\right)}}}-{1}={0}$$
Set the first factor equal to 0 and solve.
$$\displaystyle{x}={2}\pi{n},\pi+{2}\pi{n}$$ for any integer n
Set the next factor equal to 0 and solve.
$$\displaystyle{x}={\frac{{\pi}}{{{3}}}}+{2}\pi{n}$$, for any integer n
The final solution is all the values that make $$\displaystyle{\sin{{\left({x}\right)}}}{\left({2}{\cos{{\left({x}\right)}}}-{1}\right)}={0}$$ true.
$$\displaystyle{x}={2}\pi{n},\pi+{2}\pi{n},{\frac{{\pi}}{{{3}}}}+{2}\pi{n},{\frac{{{5}\pi}}{{{3}}}}+{2}\pi{n}$$, for any integer n
The finall solution is all the values that make $$\displaystyle{\sin{{\left({x}\right)}}}{\left({2}{\cos{{\left({x}\right)}}}-{1}\right)}={0}$$ true.
$$\displaystyle{x}={2}\pi{n},\pi+{2}\pi{n},{\frac{{\pi}}{{{3}}}}+{2}\pi{n},{\frac{{{5}\pi}}{{{3}}}}+{2}\pi{n}$$, for any integer n
Consolidate $$\displaystyle{2}\pi{n}$$ and $$\displaystyle\pi+{2}\pi{n}$$ to $$\displaystyle\pi{n}$$
$$\displaystyle{x}=\pi{n},{\frac{{\pi}}{{{3}}}}+{2}\pi{n},{\frac{{{5}\pi}}{{{3}}}}+{2}\pi{n}$$ for any integer n
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