\[~\left( \mathbf{v} \right)~cos\text{ }x\text{ }+\text{ }sin\text{ }x\text{ }=\text{ }cos\text{ }2x\text{ }+\text{ }sin\text{ }2x\]

Or,

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

upon rearranging we get,

\[cos\text{ }x\text{ }\text{ }cos\text{ }2x\text{ }=\text{ }sin\text{ }2x\text{ }\text{ }sin\text{ }x\]

By using the formula,

\[sin\text{ }A\text{ }\text{ }sin\text{ }B\text{ }=\text{ }2\text{ }cos\text{ }\left( A+B \right)/2\text{ }sin\text{ }\left( A-B \right)/2\]

\[cos\text{ }A\text{ }\text{ }cos\text{ }B\text{ }=\text{ }\text{ }2\text{ }sin\text{ }\left( A+B \right)/2\text{ }sin\text{ }\left( A-B \right)/2\]

So,

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

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

Or,

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

So,

\[Sin\text{ }x/2\text{ }=\text{ }0\text{ }or\text{ }sin\text{ }3x/2\text{ }=\text{ }cos\text{ }3x/2\]

\[Sin\text{ }x/2\text{ }=\text{ }sin\text{ }m\pi \]

or

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

\[Sin\text{ }x/2\text{ }=\text{ }sin\text{ }m\pi \]

Or

\[~tan\text{ }3x/2\text{ }=\text{ }1\]

\[Sin\text{ }x/2\text{ }=\text{ }sin\text{ }m\pi \]

or

\[tan\text{ }3x/2\text{ }=\text{ }tan\text{ }\pi /4\]

\[x/2\text{ }=\text{ }m\pi \]

or

\[~3x/2\text{ }=\text{ }n\pi \text{ }+\text{ }\pi /4\]

\[x\text{ }=\text{ }2m\pi \text{ }or\text{ }x\text{ }=\text{ }2n\pi /3\text{ }+\text{ }\pi /6\]

∴ the general solution is

\[x\text{ }=\text{ }2m\pi \text{ }or\text{ }2n\pi /3\text{ }+\text{ }\pi /6,\]

where m, n ϵ Z.

\[\left( \mathbf{vi} \right)~sin\text{ }x\text{ }+\text{ }sin\text{ }2x\text{ }+\text{ }sin\text{ }3x\text{ }=\text{ }0\]

Or,

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

using transformation formula,

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

By using the formula,

\[sin\text{ }A\text{ }+\text{ }sin\text{ }B\text{ }=\text{ }2\text{ }sin\text{ }\left( A+B \right)/2\text{ }cos\text{ }\left( A-B \right)/2\]

So,

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

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

Or,

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

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

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

Or,

\[Sin\text{ }2x\text{ }=\text{ }sin\text{ }0\text{ }or\text{ }cos\text{ }x\text{ }=\text{ }-1/2\]

\[Sin\text{ }2x\text{ }=\text{ }sin\text{ }0\text{ }or\text{ }cos\text{ }x\text{ }=\text{ }cos\text{ }\left( \pi \text{ }\text{ }\pi /3 \right)\]

Or,

\[Sin\text{ }2x\text{ }=\text{ }sin\text{ }0\text{ }or\text{ }cos\text{ }x\text{ }=\text{ }cos\text{ }2\pi /3\]

\[2x\text{ }=\text{ }n\pi \text{ }or\text{ }x\text{ }=\text{ }2m\pi \text{ }\pm \text{ }2\pi /3\]

Or,

\[x\text{ }=\text{ }n\pi /2\text{ }or\text{ }x\text{ }=\text{ }2m\pi \text{ }\pm \text{ }2\pi /3\]

∴ the general solution is

\[x\text{ }=\text{ }n\pi /2\text{ }or\text{ }2m\pi \text{ }\pm \text{ }2\pi /3,\]

where m, n ϵ Z.