Sum of Sines Trigonometric Equation

We can solve the equation
\[sin \theta + sin 4 \theta =sin 2 \theta n+ sin 3 \theta\]

using the identities
\[sin P +sin Q=2 sin (\frac{P+Q}{2}) cos (\frac{P+Q}{2})\]
  (1) and  
\[cos P-cos Q=2 sin (\frac{P+Q}{2})sin (\frac{P-Q}{2})\]
  (2).
using (1) on both sides of the equation gives
\[2sin \frac{5 \theta}{2} cos \frac{3 \theta}{2} = 2 sin \frac{5 \theta}{2} cos \frac{\theta}{2}\]

Subtracting  
\[2 sin \frac{5 \theta}{2} sin \frac{\theta}{2}\]
  from both sides and factorising gives
\[2sin \frac{5 \theta}{2}( cos \frac{3 \theta}{2} - cos \frac{\theta}{2})=0\]

Using identity (2) with the bracketed term gives
\[2sin \frac{5 \theta}{2}(-2sin \theta sin \frac{\theta}{2})=0\]

Hence  
\[\frac{5 \theta}{2} = n \rightarrow \theta = \frac{2n \pi}{5}\]
  or  
\[\theta = n \pi\]
  or  
\[\frac{\theta}{2} = n \pi \rightarrow \theta = 2n \pi\]
.

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