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Find the general solution of \[sinx + sin3x + sin5x = 0\].

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Last updated date: 13th Jul 2024
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Answer
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Hint – Use the formula \[\sin a + \sin b = 2\sin \left( {\dfrac{{a + b}}{2}} \right)\cos \left( {\dfrac{{a - b}}{2}} \right)\].

We have ,
\[
  sinx + sin3x + sin5x = 0 \\
  (sinx + sin5x) + sin3x = 0 \\
\]
We know ,
\[\sin a + \sin b = 2\sin \left( {\dfrac{{a + b}}{2}} \right)\cos \left( {\dfrac{{a - b}}{2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\]
Therefore,
\[sinx + sin5x = 2\sin \left( {\dfrac{{6x}}{2}} \right)\cos \left( {\dfrac{{4x}}{2}} \right) = 2\sin \left( {3x} \right)\cos \left( {2x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)\,\,\,\,\] [From (1)]
\[2\sin \left( {3x} \right)\cos \left( {2x} \right) + \sin 3x = 0\,\,\,\,\,\,\,\] [From (2)]
\[sin3x(2cos2x + 1) = 0\]
Either \[{\text{ }}sin3x = 0\;\] or \[2cos2x + 1 = 0\]
i.e. \[sin3x = 0\;\,\,\,\,{\text{or }}\;cos2x = \dfrac{{ - 1}}{2}\]
\[3x = n\pi ,\,\,\,n \in Z\;\,\,\,\,\,{\text{or}}\,\,\,\,\;2x = 2m\pi \pm \dfrac{{2\pi }}{3}\,\,\,\,\;{\text{where}}\;m \in Z\]
Hence, \[x = \dfrac{{n\pi }}{3}\;\,\,{\text{or}}\,\,\,\,\;x = m\pi \pm \dfrac{\pi }{3},{\text{ where\; }}n,m \in Z\].

Note – In these types of questions of finding general solutions, always try to simplify with the help of trigonometric formulas such that all terms on both sides are single or multiplied with each other . Then equate and then use quadrant rule in trigonometry to get the general solutions.