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How to find the max value of 12 sin x - 9 Sin^2 x.
12 sin (x)- 9 sin^2 (x) can be written as below by completing the squares method.

4 - ( 9 sin^2 (x) - 12 sin(x) + 4 )

4 - ( 3 sin (x) - 2 )^2.

4 - ( z) ^2

Now, the above expression will have max value only when z = 0.

So, the max value is 4 and occurs when Sin x = 2/3.

We can also  solve this by differentiation as given below.

As we know, the function to have max or min value its derivative should be 0.

 So, derivative of  12 sin (x)- 9 sin^2 (x) =  12 cos (x) - 2. 9. Sin (x) . Cos (x)

Equating to 0, we get  f'(x) = 12 cos (x) - 2. 9. Sin (x) . Cos (x) = 0

Simplifying,   Sin (x)  = 12/18 = 2/3 or Cos (x) = 0.

We know that to have max value f" (x) should be negative, which is satisfied by sin(x) = 2/3.

The max value is obtained by substituting Sin (x) = 2/3 in the function

that is  12 . 2/3 - 9. (2/3)^2 =  8 - 4 = 4.

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