The problem is solved here using "completing the square".
First note that we may render the given expression as:
9 sec^2 x+16cos^2 x = a^2+b^29sec2x+16cos2x=a2+b2
a = 3 sec x = 3/cos xa=3secx=3cosx
b = 4 cos xb=4cosx
Now recall the identity
(a\pm b)^2=a^2\pm 2ab+b^2(a±b)2=a2±2ab+b2
Select the minus sign and put in the values for aa and bb given above. Then 2ab2ab turns out to be a constant, 2424, and we have:
(3 sec x -4 cos x)^2=9 sec^2 x-24+16cos^2 x(3secx−4cosx)2=9sec2x−24+16cos2x
9 sec^2 x+16cos^2 x=color(blue)((3 sec x -4 cos x)^2)+249sec2x+16cos2x=(3secx−4cosx)2+24
The blue term, a squared quantity derived from the original expression, is called a completed square. We are sure that 9 sec^2 x+16cos^2 x9sec2x+16cos2x has to be at least 2424 because the completed square is nonnegative.
But is 2424 the real minimum, or might the minimum be greater because the completed square doesn't get all the way down to zero? We find out by seeing whether there are real values of xx where the completed square hits zero, thus:
3 sec x -4 cos x=03secx−4cosx=0
3/cos x = 4 cos x3cosx=4cosx
cos^2 x = 3/4cos2x=34
For real values of xx, cos^2 xcos2x can take on any value between 00 and 11. Thus we are sure that the completed square hits zero and the minimum of the given function is exactly 2424.