How Would You Solve $5\cos 6x+6=9$? 
Question: How would you solve this sinusoidal equation:

Solve $5\cos(6x)+6=9$. Assume $n$ is an integer and the answers are in degrees.
    
    
*
    
*$-8.86+n\cdot 60$
    
*$-3.54+n\cdot 60$
    
*$3.54+n\cdot 60$
    
*$8.86+n\cdot 60$
    
*$15.13+ n\cdot 360$
    
*$126.87+n\cdot 360$


I'm sort of new to this. But I have tried to isolate the trigonometric parts, and I get$$\cos(6x)=\frac 35\tag{1}$$
But after this, I'm not sure what to do. Do I take the $\arccos$ of both sides? If so, what will $\arccos\frac 35$ evaluate to? I don't think it's going to be a "perfect" number such as $\dfrac \pi 3$.
 A: I don't believe there is a "nice" way to do it. Anyway the multiple choices all being terminating decimals should give you that hint.
Starting from $\cos(6x)=\dfrac 35$, WolframAlpha gives $8.86^{\circ}$ as a solution (like you said just take $\arccos$ of both sides and divide by $6$).
Now you know if $\cos (6 \cdot 8.86) = \dfrac 35$, then $\cos (6 \cdot 8.86 + 360n) = \dfrac 35$, so  $\cos [6(8.86 + 60n)] = \dfrac 35$
Therefore $x = 8.86 + 60n$ is the solution.
If you really want to do it calculator free you can, but I wouldn't reccoment it.
EDIT:
Since cosine is an even function (i.e. $\cos (-\theta) = \cos \theta )$, another family of solutions is $-8.86 - 60n$, or just $-8.86 + 60n$. So the question has two answers.
A: $\cos 6x=3/5\iff |\cos 3x|=(\sqrt {1+\cos 6x})/2=(\sqrt {8/5})/2=\sqrt  {2/5}.$
$|\cos 3x|=\sqrt {2/5}\iff |4\cos^3x-3\cos x|=\sqrt {2/5}\iff (4\cos^3x-3\cos x)^2=4/25.$
Let $y=\cos^2 x.$ Then $\cos 6x=3/5\iff 16y^3-24y^2+9y-4/25=0.$ The cubic formula is on this website (remember to scroll down to the bottom of the page).
A: Using the cosine law you need to set the values of b and c (we are just going to have 1 variable because b and c will be equal, so we will only use b.)(A would be 6x)
$$b = (180 - 6x)/2$$
and so the formula would be (if b and c are equal): 
$$(2(b^2)-36(x^2))/2(b^2)$$
and according to wolfram-alpha, this cancels out to: 
$$1-(36(x^2)/2(b^2))$$
and inputting the value for b (using wolfram-alpha again) we get : 
$$9x^2-540x+8100$$
which is kind of a quadratic equation except it doesn't equal 0. That is the formula for $$cos(6x)$$. If you want the formula for $$cos(x)$$ you divide that by 6. 
So you want that sort-of quadratic equation to equal 3/5.
Guess what? You can turn it into a quadratic equation!: 
$$9x^2-540x+8099.4=0$$ 
So using the quadratic formula we get: 
$$-540 (+ or -) \sqrt{291600-291578.4}\over 18$$
$$-540 (+ or -) \sqrt{21.6}\over 18$$
$$-540 (+ or -) 4.64758...\over 18$$
$$-30 (+ or -) 0.25819889$$
so x is either -30.25819889 or -29.74180111.
