\[\left( \mathbf{iii} \right)~sin\text{ }x\text{ }+\text{ }sin\text{ }5x\text{ }=\text{ }sin\text{ }3x\]
Or,
\[sin\text{ }x\text{ }+\text{ }sin\text{ }5x\text{ }=\text{ }sin\text{ }3x\]
\[sin\text{ }x\text{ }+\text{ }sin\text{ }5x\text{ }\text{ }sin\text{ }3x\text{ }=\text{ }0\]
we shall rearrange and use transformation formula
\[\text{ }sin\text{ }3x\text{ }+\text{ }sin\text{ }x\text{ }+\text{ }sin\text{ }5x\text{ }=\text{ }0\]
Or,
\[\text{ }sin\text{ }3x\text{ }+\text{ }\left( sin\text{ }5x\text{ }+\text{ }sin\text{ }x \right)\text{ }=\text{ }0\]
By using the formula,
\[sin\text{ }A\text{ }+\text{ }sin\text{ }B\text{ }=\text{ }2\text{ }sin\text{ }\left( A+B \right)/2\text{ }cos\text{ }\left( A-B \right)/2\]
\[\text{ }sin\text{ }3x\text{ }+\text{ }2\text{ }sin\text{ }\left( 5x+x \right)/2\text{ }cos\text{ }\left( 5x-x \right)/2\text{ }=\text{ }0\]
\[2sin\text{ }3x\text{ }cos\text{ }2x\text{ }\text{ }sin\text{ }3x\text{ }=\text{ }0\]
Or,
\[sin\text{ }3x\text{ }\left( \text{ }2cos\text{ }2x\text{ }\text{ }1 \right)\text{ }=\text{ }0\]
\[sin\text{ }3x\text{ }=\text{ }sin\text{ }0\text{ }or\text{ }cos\text{ }2x\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]
or,
\[sin\text{ }3x\text{ }=\text{ }sin\text{ }0\text{ }or\text{ }cos\text{ }2x\text{ }=\text{ }cos\text{ }\pi /3\]
or,
\[3x\text{ }=\text{ }n\pi \text{ }or\text{ }2x\text{ }=\text{ }2m\pi \text{ }\pm \text{ }\pi /3\]
\[x\text{ }=\text{ }n\pi /3\text{ }or\text{ }x\text{ }=\text{ }m\pi \text{ }\pm \text{ }\pi /6\]
∴ the general solution is
\[x\text{ }=\text{ }n\pi /3\text{ }or\text{ }m\pi \text{ }\pm \text{ }\pi /6,\]
where m, n ϵ Z.
\[\left( \mathbf{iv} \right)~cos\text{ }x\text{ }cos\text{ }2x\text{ }cos\text{ }3x\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\]
Or,
\[cos\text{ }x\text{ }cos\text{ }2x\text{ }cos\text{ }3x\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\]
\[4\text{ }cos\text{ }x\text{ }cos\text{ }2x\text{ }cos\text{ }3x\text{ }\text{ }1\text{ }=\text{ }0\]
By using the formula,
\[2\text{ }cos\text{ }A\text{ }cos\text{ }B\text{ }=\text{ }cos\text{ }\left( A\text{ }+\text{ }B \right)\text{ }+\text{ }cos\text{ }\left( A\text{ }\text{ }B \right)\]
\[2\left( 2cos\text{ }x\text{ }cos\text{ }3x \right)\text{ }cos\text{ }2x\text{ }\text{ }1\text{ }=\text{ }0\]
Or,
\[2\left( cos\text{ }4x\text{ }+\text{ }cos\text{ }2x \right)\text{ }cos2x\text{ }\text{ }1\text{ }=\text{ }0\]
\[2\left( 2co{{s}^{2}}~2x\text{ }\text{ }1\text{ }+\text{ }cos\text{ }2x \right)\text{ }cos\text{ }2x\text{ }\text{ }1\text{ }=\text{ }0\]
\[\left[ using\text{ }cos\text{ }2A\text{ }=\text{ }2co{{s}^{2}}A\text{ }\text{ }1 \right]\]
\[4co{{s}^{3}}~2x\text{ }\text{ }2cos\text{ }2x\text{ }+\text{ }2co{{s}^{2}}~2x\text{ }\text{ }1\text{ }=\text{ }0\]
\[2co{{s}^{2}}~2x\text{ }\left( 2cos\text{ }2x\text{ }+\text{ }1 \right)\text{ }-1\left( 2cos\text{ }2x\text{ }+\text{ }1 \right)\text{ }=\text{ }0\]
Or,
\[\left( 2co{{s}^{2}}~2x\text{ }\text{ }1 \right)\text{ }\left( 2\text{ }cos\text{ }2x\text{ }+\text{ }1 \right)\text{ }=\text{ }0\]
So,
\[2cos\text{ }2x\text{ }+\text{ }1\text{ }=\text{ }0\text{ }or\text{ }\left( 2co{{s}^{2}}~2x\text{ }\text{ }1 \right)\text{ }=\text{ }0\]
\[cos\text{ }2x\text{ }=\text{ }-1/2\text{ }or\text{ }cos\text{ }4x\text{ }=\text{ }0\]
\[\left[ using\text{ }cos\text{ }2\theta \text{ }=\text{ }2co{{s}^{2}}\theta \text{ }\text{ }1 \right]\]
\[cos\text{ }2x\text{ }=\text{ }cos\text{ }\left( \pi \text{ }\text{ }\pi /3 \right)\]
or
\[cos\text{ }4x\text{ }=\text{ }cos\text{ }\pi /2\]
\[cos\text{ }2x\text{ }=\text{ }cos\text{ }2\pi /3\]
or
\[cos\text{ }4x\text{ }=\text{ }cos\text{ }\pi /2\]
\[2x\text{ }=\text{ }2m\pi \text{ }\pm \text{ }2\pi /3\]
or
\[4x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /2\]
\[x\text{ }=\text{ }m\pi \text{ }\pm \text{ }\pi /3\text{ }or\text{ }x\text{ }=\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /8\]
∴ the general solution is
\[x\text{ }=\text{ }m\pi \text{ }\pm \text{ }\pi /3\text{ }or\text{ }\left( 2n\text{ }+\text{ }1 \right)\text{ }\pi /8,\]
where m, n ϵ Z.