Solve for x: sin x +sin 3x + sin 5x= 0

sinx+sin3x+sin5x=0

⇒(sin5x+sinx)+sin3x=0

⇒2sin3xcos2x+sin3x=0

⇒sin3x(2cos2x+1)=0

sin3x=0 or 2cos2x+1=0

sin3x=0 or, cos2x=−21

Now, sin3x=0⟹3x=nπ⟹x=3nπ,n∈Z

And, cos2x=−21

cos2x=cos32π

2x=2mπ±32π,m∈Z

x=mπ±3π,m∈Z

Hence, the general solution of the given equation is :

x=3nπ or, x=mπ±3π, where m,n∈Z

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