I was able to solve it till
$$y = \sqrt{(a^2 + b^2)}\sin(\alpha + x) + C.$$
But I don't know how to find maxima and minima from here.
If $C = 0$ then maxima & minima equals the amplitude of the sine curve but when $C$ is non-zero then?
I need help from here onwards.
Hint:
$\sin(\alpha +x) \in [-1,1]$
$$ \therefore \quad \sqrt{a^2+b^2}\sin(\alpha+x) \in \left[ - \sqrt{a^2+b^2}, \sqrt{a^2+b^2} \right]$$
$$ \quad \therefore \sqrt{a^2+b^2}\sin(\alpha+x)+C \in \left[C- \sqrt{a^2+b^2}, C+ \sqrt{a^2+b^2}\ \right] \quad ,$$ therefore the maximum of $y$ is $_____$, and the minimum of $y$ is $_____$.
The old way:
$$y'=a\cos x-b\sin x=0\iff\tan x=\frac ab\iff x=\arctan\frac ab+k\pi.$$
Then
$$\sin\theta=\pm\frac{a}{\sqrt{a^2+b^2}},\\\cos\theta=\pm\frac{b}{\sqrt{a^2+b^2}},$$ and
$$y=\pm\frac{a^2+b^2}{\sqrt{a^2+b^2}}+c=\pm\sqrt{a^2+b^2}+c.$$
The maximum obviously corresponds to the plus sign.
