\[\left( \mathbf{ix} \right)~sin\text{ }2x\text{ }\text{ }sin\text{ }4x\text{ }+\text{ }sin\text{ }6x\text{ }=\text{ }0\]
Or,
\[sin\text{ }2x\text{ }\text{ }sin\text{ }4x\text{ }+\text{ }sin\text{ }6x\text{ }=\text{ }0\]
using transformation formula,
\[\text{ }sin\text{ }4x\text{ }+\text{ }sin\text{ }6x\text{ }+\text{ }sin\text{ }2x\text{ }=\text{ }0\]
By using the formula,
\[sin\text{ }A\text{ }+\text{ }sin\text{ }B\text{ }=\text{ }2\text{ }sin\text{ }\left( A+B \right)/2\text{ }cos\text{ }\left( A-B \right)/2\]
we get,
\[\text{ }sin\text{ }4x\text{ }+\text{ }2\text{ }sin\text{ }\left( 6x+2x \right)/2\text{ }cos\text{ }\left( 6x-2x \right)/2\text{ }=\text{ }0\]
\[\text{ }sin\text{ }4x\text{ }+\text{ }2\text{ }sin\text{ }4x\text{ }cos\text{ }2x\text{ }=\text{ }0\]
Or,
\[Sin\text{ }4x\text{ }\left( 2\text{ }cos\text{ }2x\text{ }\text{ }1 \right)\text{ }=\text{ }0\]
So,
\[Sin\text{ }4x\text{ }=\text{ }0\text{ }or\text{ }2\text{ }cos\text{ }2x\text{ }\text{ }1\text{ }=\text{ }0\]
\[Sin\text{ }4x\text{ }=\text{ }sin\text{ }0\text{ }or\text{ }cos\text{ }2x\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]
Or,
\[Sin\text{ }4x\text{ }=\text{ }sin\text{ }0\text{ }or\text{ }cos\text{ }2x\text{ }=\text{ }\pi /3\]
\[4x\text{ }=\text{ }n\pi \text{ }or\text{ }2x\text{ }=\text{ }2m\pi \text{ }\pm \text{ }\pi /3\]
Or,
\[x\text{ }=\text{ }n\pi /4\text{ }or\text{ }x\text{ }=\text{ }m\pi \text{ }\pm \text{ }\pi /6\]
∴ the general solution is
\[x\text{ }=\text{ }n\pi /4\text{ }or\text{ }m\pi \text{ }\pm \text{ }\pi /6,\]
where m, n ϵ Z.