Hello Friend,here is the solution for your question
<span>so the given function is </span> 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 :- <span>(cosx=3/4)²----------------(2) is to it's minimum value then the whole equation </span> <span>i.e. = √[1/8-3(cosx=3/4)²] will be maximum and vice versa </span>
And we know that cosx-3/4 will be minimum if cosx=3/4 <span>therefore put this in (1) we get </span> (cosx=3/4)²=0 [ cosx=3/4] <span>hence the minimum value of the quantity (cosx=3/4)² is 0 </span>
<span>put this in equation (1) </span> 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)
<span>this is the maximum value now to find the minimum value </span>
<span>since this is function of root so the value of y will always be ≥0 </span>
<span>hence the minimum value of the function y is 0 </span>
<span>Therefore, the range of function </span>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._
