SOLUTION: The equation y=3cosA+4sinA has a real solution if,

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Question 1088218: The equation y=3cosA+4sinA has a real solution if,
Answer by ikleyn(52880) About Me  (Show Source):
You can put this solution on YOUR website!
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If  y = 3cosA + 4sinA,  then  (divide both sides by 5. You will get)


y%2F5 = %283%2F5%29%2Acos%28A%29+%2B+%284%2F5%29%2Asin%28A%29.


Since %283%2F5%29%5E2+%2B+%284%2F5%29%5E2 = 1, there exist the angle alpha such as 


    sin%28alpha%29 = 3%2F5,  cos%28alpha%29 = 4%2F5.


Then we have 

y%2F5 = sin%28alpha%29%2Acos%28A%29 + cos%28alpha%29%2Asin%28A%29 = sin%28alpha+%2B+A%29.

Since  -1 <= sin(alpha +A) <= 1, then  -1 < y%2F5 <= 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].