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

Question

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 1

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

Answer 2

You already have an answer that helps with the trigonometry. I am responding to this direct question.

Am I allowed to use the identity ... because the 9 is multiplying the cos only, so that I'm not allowed to use this identity?

You are right to be worried. You can't, because that violates the distributive law. You know perfectly well that $$ a(b+c) = ab + ac \ne ab + c . $$

You just forgot that elementary fact while focusing on the trigonometry.

Answer 3

Since $\cos x=0$ is not a solution, we have that

$$\sin x = 3\cos x\implies \frac{\sin x}{\cos x} = 3\frac{\cos x}{\cos x}\implies \tan x = 3$$

and therefore

$$x=\arctan 3 +k\pi$$