Finding $x$ values of $\cos 6x + 1 = \frac{3}{2} + \frac {1}{2} \cos 3x $ 
Solve $$\cos 6x + 1 = \frac{3}{2} + \frac {1}{2} \cos 3x $$ for $0^\circ<x<120^\circ$

I simplify it to 
$$2 \cos 6x + 2 = 3 + \cos 3x $$
There is $\cos 6x $ and $\cos 3x$ how do I merge them together? To solve the equation from $0^\circ$ to $120^\circ$? 

Here's what I got...(EDIT: $3 \cos 3x$ is incorrect in the third line; the correct quantity is $\cos 3x$.).
$$\cos 6x + 1 = \frac {3}{2} + \frac {1}{2} \cos 3x$$
Multiplying by $2$ we get
$$2 \cos 6x + 2 = 3 + \cos 3x$$
Swapping constants and bringing over $\cos 3x$ we then get
$$2 \cos 6x - \cos 3x = 1$$
Using the double angle identity $\cos 6x = 2 \cos^2 3x - 1$ we get
$$4 \cos^2 3x - 2 -  \cos 3x = 1$$
Adding 2 to each side gives us
$$4 \cos^2 3x -  \cos 3x = 3$$
Now we can factor out $\cos 3x$ on the left to get
$$\cos 3x (4 \cos 3x - 1) = 3$$
We now have two equations.  Since $\cos 3x = 3$ has no solution, we concentrate on $4 \cos 3x - 1 = 3$; this yields
$$4 \cos 3x = 4$$ or $$\cos 3x = 1$$
Taking the inverse cosine of both sides we get
$$3x = 0 + 360\times k$$ (k in degrees)
and then dividing by 3 we get
$$x = 0 + 120 \times k$$
For the condition $x \in (0, 120)$ we arrive at the answer, $$\bbox [yellow] {x=120}$$.
CHECK: $\cos 6(120) + 1 \rightarrow 1 + 1 \rightarrow 2$ and $\frac {3}{2} + \frac {1}{2} \cos 360 \rightarrow \frac {3}{2} + \frac {1}{2} \rightarrow \frac {4}{2} \rightarrow 2 \checkmark$
Alternative answer (as suggested by Lab Bhatterchajee):
Instead of factoring out as above, substitute $u = \cos 3x$ and subtract $3$ from both sides gives us
$$4u^2 - u + 3 = 0$$
Using the quadratic equation $x = \frac {-b \pm \sqrt {b^2-4ac}}{2a}$ we have
$$u = \frac {-1 \pm \sqrt {1^2-4(4)(-3)}}{(2)(4)}$$, which leaves us with 
$$u = \frac {1 \pm 7}{8}$$
Bringing back $u = \cos 3x$ we have two solutions:
$$\cos 3x = -\frac {3}{4}, 1$$
Since we are looking for values from $0$ to $120$, we discard the negative value and continue on from $\cos 3x = 1$ from before - yielding the same answer as above, avoiding an extraneous solution.
