What is the range of function y=√(-2cos^2x+3cosx-1)

Hello Friend,here is the solution for your question

so the given function is 
y= √(-2cos²x+3cosx-1)
 i.e = √[-2(cos²x-3/2+1/2)]
i.e = √[-2(cosx-3/4)²-9/16+1/2]
i.e. = √[-2(cos-3/4)²-1/16]
i.e. = √[1/8-3(cosx=3/4)²](1)

Now here  in this equation is this quantity :-
(cosx=3/4)²(2)   is to it's minimum value then the whole equation 
i.e. = √[1/8-3(cosx=3/4)²] will be maximum and vice versa 

And we know that cosx-3/4 will be minimum if cosx=3/4
therefore put this in (1) we get 
(cosx=3/4)²=0    [ cosx=3/4]
hence the minimum value of the quantity (cosx=3/4)² is 0 

put this in equation (1) 
we get ,
i.e. = √[1/8-3(cosx=3/4)²]
   =√[1/8-3(0)]        [ because minimum value of of the quantity (cosx=3/4)² is 0 ]
     =√1/8
      =1/(2√2)

this is the maximum value now to find the minimum value 

since this is function of root so the value of y will always be ≥0 

hence the minimum value of the function y is 0 

Therefore, the range of function y is [0,1/(2√2)]

__Well,I have explained explained each and every step,do tell me if you don't understand any step._

How much is 32 out of 40 written as a percentage? convert fraction (ratio) 32 / 40 answer:   80%