If $0\leq a \leq 3; 0\leq b \leq 3$ and the equation $x^2 +4+3 cos(ax+b)=2x$ has at least one solution , then find the value of a+b Problem : If $0\leq a \leq 3; 0\leq b \leq 3$ and the equation $x^2 +4+3 cos(ax+b)=2x$ has at least one solution , then find the value of a+b. 
Solution : 
We can write  the given equation : $x^2 +4+3 cos(ax+b)=2x$ as $x^2-2x+4 =-3cos(ax+b)$
Since the L.H.S.  of this problem is always $+ve$ $\forall x \in \mathbb{R}$ , and we know that $ -1 \leq cosx \leq 1$ 
But I am not getting any clue how to proceed further please help. 
Thanks

$$x^2-2x+4 =-3\cos(ax+b)$$
$$(x-1)^2 +3 =-3\cos(ax+b)$$
$$\frac{(x-1)^2 +3}{-3} =\cos(ax+b)$$
how $-1\leq\cos{(ax+b)}\leq 1$, then $x=1$
$$\cos(a+b)=-1$$
$$\cos(a+b)=-1$$
$$a+b=\pi$$
because $a,b\leq3$
