If  y = 3cosA + 4sinA,  then  (divide both sides by 5. You will get)


 = .


Since  = 1, there exist the angle  such as 


     = ,   = .


Then we have 

 =  +  = .

Since  -1 <= sin(alpha +A) <= 1, then  -1 <  <= 1,   or

- 5 <= y <= 5.


From this point, you can easily convince yourself and prove that The equation y = 3cosA + 4sinA  has a real solution 
if and only if  y  belongs to the segment [-5,5].


Answer.  The equation y = 3cosA + 4sinA  has a real solution if  y  is in the segment [-5,5].