trigonometry - Can I use $cos(2x)=cos^2(x)-sin^2(x)$ to rewrite $9cos^2(x)-sin^2(x)$ as $9cos(2x)$?

I need to solve for $x$ in

$$\sin(x) = 3\cos(x)$$

So I did the following:
$$\begin{align}
\sin(x) &= 3\cos(x) \\[4pt]
\sin^2(x) &= 9\cos^2(x) &\text{(squaring both sides)}\\[4pt]
0 &= 9\cos^2(x)-\sin^2(x) &\text{(subtracting $\sin^2(x)$)}
\end{align}$$

My question is: Am I allowed to use the identity

$$\cos(2x) = \cos^2(x) - \sin^2(x)$$
in the equation to make it
$$0 = 9\cos^2(x)-\sin^2(x)\quad\to\quad 0 = 9\cos(2x)$$

or is that the case that
$$9\cos(2x)\neq 9\cos^2(x)-\sin^2(x)$$
because the $9$ is multiplying the $\cos$ only, so that I'm not allowed to use this identity?

Answer

Hint: once squared, you can use the identity:
$$\sin^2(x)+\cos^2(x)=1$$

To obtain
$$\sin^2(x)=9\cos^2(x)=9\cdot(1-\sin^2(x)).$$
From here you get
$$\sin^2(x)=9/10$$
And you can finish the calculation...

There is no need to use the double-angle identity