i have confusion in finding range of sin^4x + cos^2x.
I have replace cos^2x as sin^2x -1 and let sin^2x as t.
It becomes then t^2-t+1. Now how can i proceed?
Well I would say the range of t is (0.75, $\displaystyle \infty$)
And as Opalg already suggested the range between 0 and 1 then the original functions range is the intersection of the 2.
I.e $\displaystyle \displaystyle [0,1] \cap \left[\frac{3}{4}, \infty\right) = \dots$
You have to think about what happens as t goes from 0 to 1. As t goes from 0 to 1/2, $\displaystyle (t-\tfrac12)^2$ goes from 1/4 down to 0. And as t goes from 1/2 to 1, $\displaystyle (t-\tfrac12)^2$ goes from 0 back up to 1/4. So the range of $\displaystyle (t-\tfrac12)^2$ will be the interval [0,1/4]. It follows that the range of $\displaystyle (t-\tfrac12)^2 + \tfrac34$ is the interval [3/4,1].