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Question and Answer

Find the general solution for  the following equation: \( sinx+sin3x+sin5x=0\)

Answer

\(x\in \dfrac{n\pi }{3}\cup (n\pi \pm \dfrac{\pi }{3})\)
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Solution

\( sinx+sin3x+sin5x=0\)
\(\Rightarrow 2sin(\dfrac{5x+x}{2})cos(\dfrac{5x-x}{2})+sin3x=0\)   [ \(sinA+sinB=2sin(\dfrac{A+B}{2})cos(\dfrac{A-B}{2})\)]
\(\Rightarrow 2sin3xcos2x+sin3x=0\)
\(\Rightarrow sin3x(2cos2x+1)=0\)
\(\Rightarrow sin3x=0\)
\(\Rightarrow sin3x=sin\pi\)
Use general solution of \(sin\theta =sin\alpha\)
\(\theta =n\pi +(-1){}^{n}\alpha ,\;\;\;n\in Z\)
\(\Rightarrow 3x=n\pi +(-1){}^{n}\pi\)
\(\Rightarrow x=\dfrac{n\pi }{3}+(-1){}^{n}\dfrac{\pi }{3}\)....(i)
or \(cos2x=\dfrac{-1}{2}\)
\(cos2x=cos(\dfrac{2\pi }{3})\)
\(2x=2n\pi \pm \dfrac{2\pi }{3}\)  [ \(cos\theta =cos\alpha \;then\;\theta =2n\pi \pm \alpha\)]
\(\Rightarrow x=n\pi \pm \dfrac{\pi }{3}\)....(ii)
\(x\in \dfrac{n\pi }{3}\cup (n\pi \pm \dfrac{\pi }{3})\)
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