Question: Please help in Solving the Trigonometric Equation:

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

Answer: 
$$
\cos 3x=\cos x \cos x - \sin x \sin x= \cos 2x,
$$
which is equivalent to
$$
3x\pm 2x = 2k\pi
$$
for some integer $k$.
$$\cos^2 x-\sin^2 x = \cos 3x$$

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 $\displaystyle x= \frac{2n\pi}{5}\;,$