GIVEN:

\[sin\text{ }x\text{ }+\text{ }sin\text{ }3x\text{ }+\text{ }sin\text{ }5x\text{ }=\text{ }0\]

OR,

\[\left( sin\text{ }x\text{ }+\text{ }sin\text{ }5x \right)\text{ }+\text{ }sin\text{ }3x\text{ }=\text{ }0\]

BY FORMULA,

OR,

\[2\text{ }sin\text{ }3x\text{ }cos\text{ }\left( -\text{ }2x \right)\text{ }+\text{ }sin\text{ }3x\text{ }=\text{ }0\]

OR,

\[2\text{ }sin\text{ }3x\text{ }cos\text{ }2x\text{ }+\text{ }sin\text{ }3x\text{ }=\text{ }0\]

COMMON TERMS ARE TAKEN OUT

NOW,

\[sin\text{ }3x\text{ }\left( 2\text{ }cos\text{ }2x\text{ }+\text{ }1 \right)\text{ }=\text{ }0\]

Here

\[sin\text{ }3x\text{ }=\text{ }0\text{ }or\text{ }2\text{ }cos\text{ }2x\text{ }+\text{ }1\text{ }=\text{ }0\]

In the event that wrongdoing \[3x\text{ }=\text{ }0\]

\[3x\text{ }=\text{ }n\pi ,\text{ }where\text{ }n\in Z\]

We get

\[x\text{ }=\text{ }n\pi /3,\text{ }where\text{ }n\in Z\]

In the event that \[2\text{ }cos\text{ }2x\text{ }+\text{ }1\text{ }=\text{ }0\]

\[cos\text{ }2x\text{ }=\text{ }\text{ }1/2\]

By additional rearrangements

\[=\text{ }\text{ }cos\text{ }\pi /3\]

\[=\text{ }cos\text{ }\left( \pi \text{ }\text{ }\pi /3 \right)\]

So we get

\[cos\text{ }2x\text{ }=\text{ }cos\text{ }2\pi /3\]

Here