\[~\left( \mathbf{vii} \right)~cos\text{ }4x\text{ }=\text{ }cos\text{ }2x\]
Or,
\[cos\text{ }4x\text{ }=\text{ }cos\text{ }2x\]
\[4x\text{ }=\text{ }2n\pi \text{ }\pm \text{ }2x\]
So,
\[4x\text{ }=\text{ }2n\pi \text{ }+\text{ }2x\text{ }\left[ or \right]\text{ }4x\text{ }=\text{ }2n\pi \text{ }\text{ }2x\]
\[2x\text{ }=\text{ }2n\pi \]
[or]
\[6x\text{ }=\text{ }2n\pi \]
\[x\text{ }=\text{ }n\pi \]
[or]
\[x\text{ }=\text{ }n\pi /3\]
∴ the general solution is
\[x\text{ }=\text{ }n\pi \text{ }\left[ or \right]\text{ }n\pi /3\]
, where n ϵ Z.