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
If we complete the square we get
$$x^2 -2x +4 = (x-1)^2 +3 \geq 3$$
Further we note that $-3\cos(ax+b) \leq 3$. So it must be that we have $x^2-2x+4 = 3 = -3\cos(ax+b)$. Further, from the completing the square formula, we see that in fact we need $x=1$.
So it reduces down to the problem of solving $-3\cos(a+b) = 3$ or $\cos(a+b) = -1$
Hence $a+b = (2n+1)\pi$ for any integer $n$.
EDIT: Since $0 \leq a \leq 3$, $0 \leq b \leq 3$, we get $0 \leq a+b \leq 6$. Hence the only viable solution would be $\pi$.
$$x^2-2x+4 =-3\cos(ax+b)$$
$$(x-1)^2 +3 =-3\cos(ax+b)$$
$$\frac{(x-1)^2 +3}{-3} =\cos(ax+b)$$
how $-1\leq\cos{(ax+b)}\leq 1$, then $x=1$
$$\cos(a+b)=-1$$ $$\cos(a+b)=-1$$ $$a+b=\pi$$ because $a,b\leq3$
