If $x^2 + 4 + 3\sin(ax+b)-2x=0$ has at least one real solution, ... 
If the equation 
  $$x^2 + 4 + 3\sin(ax+b)-2x=0$$
  has at least one real solution, where $a$ and $b$ belong to $(0,2\pi)$, then what can a possible value of $(a+b)$ be?

Can anyone say what exactly should be done in the question?
Value of $\sin (ax + b)$ could be found, but how can we know about the value of $(a+b)$?

$x^2 + 4 + 3sin(ax+b)-2x=0$ can be written as $(x-1)^2+3= -3sin(ax+b)$ now you can observe that left hand side will always be greater than or equal to 3, while right hand side will always be less than or equal to 3.
hence for consistent system both sides will be equal to 3 .that implies $x=1$ further $sin(ax+b)=-1$ putting x in this will give $sin(a+b)=-1$
under given interval (0,2$\pi$)    $a+b$ can be equal to ${\frac{3\pi}{2}}$ and ${\frac{7\pi}{2}}$ ;as $a+b$ will lie between $0$ and $4\pi$
