How do you solve \sin2x=\sin x?

Salvatore Boone 2021-12-24 Answered
How do you solve \(\displaystyle{\sin{{2}}}{x}={\sin{{x}}}\)?

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Expert Answer

Philip Williams
Answered 2021-12-25 Author has 3856 answers
\(\displaystyle{\sin{{2}}}{x}={\sin{{x}}}\)
\(\displaystyle{2}{\sin{{x}}}{\cos{{x}}}={\sin{{x}}}\)
\(\displaystyle{2}{\sin{{x}}}{\cos{{x}}}-{\sin{{x}}}={0}\)
\(\displaystyle{\sin{{x}}}{\left({2}{\cos{{x}}}-{1}\right)}={0}\)
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
Answered 2021-12-26 Author has 1267 answers
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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user_27qwe
Answered 2021-12-30 Author has 9558 answers
Very good answer
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