Trig Q.

Let y = 16cos^2 x - 24cos x + 9
Factorize it
y = (4cos x - 3)^2 = 0

Find the minimum value of y
Reconsider y = w^2
Square of a number, w (either positive number or negative number) always gives a positive value.
For minimum value of y, w = 0
w = (4 cos x - 3) = 0
For minimum value of y, cos x = ¾
y = (4 cos x - 3)^2 = 0
y = (4 (¾) - 3)^2 = 0
The minimum value of 16 cos^2 x - 24 cos x + 9 is 0

For maximum value of y
y = (4 cos x - 3)^2
(4 cos x) must be negative number and the absolute value of (4cos x) must be as large as possible.
[ -1 <= cos x <= +1]
This can occur if cos x = -1
y = (4 cos x - 3)^2
y = (4 (-1) - 3)^2
y = (-4 - 3)^2
y = (-7)^2
y = + 49

The minimum value of 16 cos^2 x - 24cos x + 9 is 0 (Answer)
The maximum value of 16 cos^2 x - 24cos x + 9 is +49 (Answer)

Find the maximum and minimum value of 16Cos^2x－24Cos x+9
Sol
16Cos^2 x－24Cos x+9
=16(Cos^2 x－1.5Cos x)+9
=16(Cos^2 x－2*Cosx*0.75+0.75^2)+9－16*0.75^2
=16(Cosx－0.75)^2
－1<=Cosx<=1
－1.75<=Cosx－0.75<=0.25
0<=(Cosx－0.75)^2<=1.75^2
0<=16(Cosx－0.75)^2<=16*1.75^2
0<=16Cos^2x－24Cos x+9<=49