Solve $\sin \left( {4\theta } \right) = \cos \left( {2\theta } \right)$.

Hint:Using some basic trigonometric identities like:
$
\sin \left( {2\theta } \right) = 2\sin \left( \theta \right)\cos \left( \theta \right) \\
\cos \left( {2\theta } \right) = {\cos ^2}\left( \theta \right)-{\sin ^2}\left( \theta \right) = 1-2{\text{
}}{\sin ^2}\left( \theta \right) = 2{\text{ }}{\cos ^2}\left( \theta \right)-1 \\
$
Such that in order to solve and simplify the given expression we have to use the above identities and express our given expression in that form and thereby simplify and solve for ${\theta _{}}$.

Complete step by step solution:
Given
$\sin \left( {4\theta } \right) = \cos \left( {2\theta } \right).............................\left( i \right)$
Here we have to basically solve for ${\theta _{}}$ using any mathematical operations and functions. In order to solve for ${\theta _{}}$ we can use the above mentioned trigonometric identities to simplify it.
Since\[\sin \left( {2\theta } \right) = 2\sin \left( \theta \right)\cos \left( \theta \right)\], using this expression to represent $\sin \left( {4\theta } \right)$, and we get:
$\sin \left( {4\theta } \right) = 2\cos \left( {2\theta } \right)\sin \left( {2\theta }
\right).....................\left( {ii} \right)$
Substituting (ii) in (i) we get:
$2\cos \left( {2\theta } \right)\sin \left( {2\theta } \right) = \cos \left( {2\theta }
\right)........................\left( {iii} \right)$
Now to solve for $\theta $ add $ - \cos \left( {2\theta } \right)$ to both the LHS and RHS, such that we get:
$2\cos \left( {2\theta } \right)\sin \left( {2\theta } \right) - \cos \left( {2\theta } \right) = 0$
On simplifying we can write:
$
\Rightarrow 2\cos \left( {2\theta } \right)\sin \left( {2\theta } \right) - \cos \left( {2\theta } \right) =
0 \\
\Rightarrow \cos \left( {2\theta } \right)\left( {\left( {2\sin 2\theta } \right) - 1} \right) = 0 \\
$
Now equating each terms to RHS which is zero, we get:
$\cos \left( {2\theta } \right) = 0\;\;{\text{and}}\;\;\left( {2\sin 2\theta } \right) - 1 = 0$
Taking the first part and the second part separately:
$
\cos \left( {2\theta } \right) = 0 \\
\Rightarrow 2\theta = {\cos ^{ - 1}}\left( 0 \right) \\
\Rightarrow 2\theta = \left( {2n + 1} \right)\dfrac{\pi }{2} \\
\Rightarrow \theta = \left( {2n + 1} \right)\dfrac{\pi }{4}......................\left( {iv} \right) \\
$
So we got one part of our solution. Now equating the second part we get:
$
\left( {2\sin 2\theta } \right) - 1 = 0 \\
\Rightarrow 2\sin 2\theta = 1 \\
\Rightarrow \sin 2\theta = \dfrac{1}{2} \\
\Rightarrow 2\theta = n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{6} \\
\Rightarrow \theta = \dfrac{{n\pi }}{2} + {\left( { - 1} \right)^n}\dfrac{\pi
}{{12}}............................\left( v \right) \\
$
So the solution for $\theta $is given by the equations (iv) and (v).
Therefore on solving $\sin \left( {4\theta } \right) = \cos \left( {2\theta } \right)$ we get:
$\theta = \left( {2n + 1} \right)\dfrac{\pi }{4}\;\;{\text{or}}\;\;\theta = \dfrac{{n\pi }}{2} + {\left( { - 1} \right)^n}\dfrac{\pi }{{12}}$

Note:
General formulas:
$
\cos \theta = 0 \\
\Rightarrow \theta = \left( {2n + 1} \right)\dfrac{\pi }{2} \\
\sin \theta = 0 \\
\Rightarrow \theta = n\pi \\
$
The above formulas are quite useful for solving and simplifying a trigonometric equation without any complications. Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.