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. $2\pi n=\frac{2\pi(5n)}5$.
As others have mentioned you should be after the roots of $\cos(3x)-\cos(2x)=0$. To continue from here you may use the identity $\cos \theta-\cos \alpha=-2\sin\frac{\theta-\alpha}{2}\sin\frac{\theta+\alpha}{2}$, i.e. 

$$\cos(3x)-\cos(2x)=-2\sin\frac{5x}{2}\sin\frac{x}{2}$$