TITLE: Quadratic Formula, nature of roots with Trigonometric Functions
QUESTION [9 upvotes]: The original problem:

If $0\le a,b\le 3$ and the equation $$x^2+4+3\cos(ax+b)=2x$$ has at least one real solution, then find the value of $a+b$

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At first, on rearranging, I got the following expression:
$$x^2-2x+(4+3\cos(ax+b))=0$$ I thought this was a quadratic in $x$, and thus from the quadratic formula(and that at least one real root exists), $D\ge 0$ ie $$4-4(4+\cos(ax+b))\ge 0$$ 
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However I'm not really sure about this. I've treated $\cos(ax+b)$ as a constant term even though the argument of the cosine includes $x$: the variable in which the quadratic expression is.
$$$$Under these circumstances, is it correct to use $3\cos(ax+b)$ as a constant? If not, how could I use the quadratic formula to find values of $x$ satisfying $$x^2-2x+(4+3\cos(ax+b))=0$$
Many thanks in anticipation!

REPLY [6 votes]: Completing the square,
$$x^2-2x+1=(x-1)^2=-3(1+\cos(ax+b)).$$
The two expressions can only be equal if they are zero, i.e. $x=1$ and $\cos(ax+b)=-1$.
Hence,
$$a+b=\pi.$$