Question: Please help in Solving the Trigonometric Equation:

$$\cos^2x - \sin^2x = \cos3x$$

Answer: 
$$\begin{align\*}
\cos^2x-\sin^2x &= \cos3x\\
\cos2x &= \cos3x\\
2x &= 2\pi n\pm 3x\\
x &= 2\pi n \text{ or } x=\frac{2\pi n}{5}
\end{align\*}$$
where $n\in\mathbb Z$.

@ThomasAndrews: The second case includes the first case. $2\pi n=\frac{2\pi(5n)}5$.
$$\cos^2 x-\sin^2 x = \cos 3x$$

Now Using the formula $\cos2x = \cos^2 x-\sin^2 x$

So $$\cos 2x = \cos 3x\Rightarrow \cos3x = \cos 2x\Rightarrow 3x = 2n\pi\pm 2x$$

Where $n\in \mathbb{Z}\;,$ above we have used the formula $$\cos x = \cos \alpha \;,$$ Then $x=2n\pi\pm \alpha\;,$ Where $n\in \mathbb{Z}$

So Solutions are $3x = 2n\pi+2x$ or $3x = 2n\pi-2x$$

So we get $x=2n\pi$ or $\displaystyle x = \frac{2n\pi}{5}\;,$ Where $n\in \mathbb{Z}$

So Final Solution is $\displaystyle x= \frac{2n\pi}{5}\;,$