# The ratio of the greatest value of 2-cosx+sin^2x to its least value is? After a point there comes a quadratic equation of whom min and max is to be found. Please help

Abhishek Singh
105 Points
7 years ago
you can convert sin2 x to 1-cos2
and the equation becomes. 3 – cos(x) - cos2x  which surely is a quadratic equation in cos(x). substitute cos(x) as t.
Now your expression becomes 3-t-t2 . Now plot 3-t-t^2. But only where  -1
so ratio is 4/3.
Hope  it helps
Ajay
209 Points
7 years ago
it can be easily solved by finding maxima and minima
the given expression translates to 3-cos x – cos2x
Differentiating and euating to 0.
sinx + 2cos x sin x = 0
sinx (1+ 2 Cos x)= 0
sinx = 0 or cosx = -1/2
cos x  =1 or cos x  = -1/2
Cos x  = 1 will give minimum value and  cos x  =-1/2 will give max.
Substituting this in expresstion 3-cos x – cos2x
Max Value = 3 + ½ -1/4 = 13/4
and Min value = 3 – 1 – 1 = 1
Hence ratio of max and min  = 13/4.
I have confirmed thing by plotting the graph of entire function.