\[~\left( \mathbf{vii} \right)~sin\text{ }x\text{ }+\text{ }sin\text{ }2x\text{ }+\text{ }sin\text{ }3x\text{ }+\text{ }sin\text{ }4x\text{ }=\text{ }0\]
Or,
\[sin\text{ }x\text{ }+\text{ }sin\text{ }2x\text{ }+\text{ }sin\text{ }3x\text{ }+\text{ }sin\text{ }4x\text{ }=\text{ }0\]
using transformation formula,
\[sin\text{ }x\text{ }+\text{ }sin\text{ }3x\text{ }+\text{ }sin\text{ }2x\text{ }+\text{ }sin\text{ }4x\text{ }=\text{ }0\]
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,
\[2\text{ }sin\text{ }\left( 3x+x \right)/2\text{ }cos\text{ }\left( 3x-x \right)/2\text{ }+\text{ }2\text{ }sin\text{ }\left( 4x+2x \right)/2\text{ }cos\text{ }\left( 4x-2x \right)/2\text{ }=\text{ }0\]
\[2\text{ }sin\text{ }2x\text{ }cos\text{ }x\text{ }+\text{ }2\text{ }sin\text{ }3x\text{ }cos\text{ }x\text{ }=\text{ }0\]
Or,
\[2cos\text{ }x\text{ }\left( sin\text{ }2x\text{ }+\text{ }sin\text{ }3x \right)\text{ }=\text{ }0\]
Again 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,
\[2cos\text{ }x\text{ }\left( 2\text{ }sin\text{ }\left( 3x+2x \right)/2\text{ }cos\text{ }\left( 3x-2x \right)/2 \right)\text{ }=\text{ }0\]
\[2cos\text{ }x\text{ }\left( 2\text{ }sin\text{ }5x/2\text{ }cos\text{ }x/2 \right)\text{ }=\text{ }0\]
Or,
\[4\text{ }cos\text{ }x\text{ }sin\text{ }5x/2\text{ }cos\text{ }x/2\text{ }=\text{ }0\]
So,
\[Cos\text{ }x\text{ }=\text{ }0\text{ }or\text{ }sin\text{ }5x/2\text{ }=\text{ }0\text{ }or\text{ }cos\text{ }x/2\text{ }=\text{ }0\]
\[Cos\text{ }x\text{ }=\text{ }cos\text{ }0\]
or
\[sin\text{ }5x/2\text{ }=\text{ }sin\text{ }0\]
or
\[cos\text{ }x/2\text{ }=\text{ }cos\text{ }0\]
\[Cos\text{ }x\text{ }=\text{ }cos\text{ }\pi /2\]
or
\[sin\text{ }5x/2\text{ }=\text{ }k\pi \]
or
\[cos\text{ }x/2\text{ }=\text{ }cos\text{ }\left( 2p\text{ }+\text{ }1 \right)\text{ }\pi /2\]
\[x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /2\]
or
\[5x/2\text{ }=\text{ }k\pi \text{ }or\text{ }x/2\text{ }=\text{ }\left( 2p\text{ }+\text{ }1 \right)\text{ }\pi /2\]
\[x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /2\]
or
\[x\text{ }=\text{ }2k\pi /5\text{ }or\text{ }x\text{ }=\text{ }\left( 2p\text{ }+\text{ }1 \right)\]
\[x\text{ }=\text{ }n\pi \text{ }+\text{ }\pi /2\]
or
\[x\text{ }=\text{ }2k\pi /5\text{ }or\text{ }x\text{ }=\text{ }\left( 2p\text{ }+\text{ }1 \right)\]
∴ general solution:
\[x\text{ }=\text{ }n\pi \text{ }+\text{ }\pi /2\]
or
\[x\text{ }=\text{ }2k\pi /5\]
or
\[x\text{ }=\text{ }\left( 2p\text{ }+\text{ }1 \right),\]
where n, k, p ϵ Z.
\[\left( \mathbf{viii} \right)~sin\text{ }3x\text{ }\text{ }sin\text{ }x\text{ }=\text{ }4\text{ }co{{s}^{2}}~x\text{ }\text{ }2\]
Or,
\[sin\text{ }3x\text{ }\text{ }sin\text{ }x\text{ }=\text{ }4\text{ }co{{s}^{2}}~x\text{ }\text{ }2\]
\[sin\text{ }3x\text{ }\text{ }sin\text{ }x\text{ }=\text{ }2\left( 2\text{ }co{{s}^{2}}~x\text{ }\text{ }1 \right)\]
or,
\[sin\text{ }3x\text{ }\text{ }sin\text{ }x\text{ }=\text{ }2\text{ }cos\text{ }2x\]
\[\left[ as,~cos\text{ }2A\text{ }=\text{ }2co{{s}^{2}}~A\text{ }\text{ }1 \right]\]
By using the formula,
\[Sin\text{ }A\text{ }\text{ }sin\text{ }B\text{ }=\text{ }2\text{ }cos\text{ }\left( A+B \right)/2\text{ }sin\text{ }\left( A-B \right)/2\]
So,
\[2\text{ }cos\text{ }\left( 3x+x \right)/2\text{ }sin\text{ }\left( 3x-x \right)/2\text{ }=\text{ }2\text{ }cos\text{ }2x\]
\[2\text{ }cos\text{ }2x\text{ }sin\text{ }x\text{ }\text{ }2\text{ }cos\text{ }2x\text{ }=\text{ }0\]
Or,
\[2\text{ }cos\text{ }2x\text{ }\left( sin\text{ }x\text{ }\text{ }1 \right)\text{ }=\text{ }0\]
Then,
\[2\text{ }cos\text{ }2x\text{ }=\text{ }0\text{ }or\text{ }sin\text{ }x\text{ }\text{ }1\text{ }=\text{ }0\]
\[Cos\text{ }2x\text{ }=\text{ }0\text{ }or\text{ }sin\text{ }x\text{ }=\text{ }1\]
Or,
\[Cos\text{ }2x\text{ }=\text{ }cos\text{ }0\text{ }or\text{ }sin\text{ }x\text{ }=\text{ }sin\text{ }1\]
\[Cos\text{ }2x\text{ }=\text{ }cos\text{ }0\text{ }or\text{ }sin\text{ }x\text{ }=\text{ }sin\text{ }\pi /2\]
Or,
\[2x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /2\]
or
\[x\text{ }=\text{ }m\pi \text{ }+\text{ }{{\left( -1 \right)}^{~m}}~\pi /2\]
\[x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /4\]
or
\[x\text{ }=\text{ }m\pi \text{ }+\text{ }{{\left( -1 \right)}^{~m}}~\pi /2\]
∴ general solution:
\[x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /4\text{ }or\text{ }m\pi \text{ }+\text{ }{{\left( -1 \right)}^{~m}}~\pi /2,\]
where m, n ϵ Z.