How do you find the interval of decrease and increase of a sin function? Given this functionf(x) = sin2x-90degreesWhat is the interval of increase and decrease and how do you find the zeros? Your help is greatly appreciated! ThanksThe function f(x) is incresing when f'(x) is >0.f'(x) = (sin2x -90)' = 2cos2x2co2x > 0 implies cos2x > 0Cos2x > 0 when 2npi-pi/2 < 2x < 2npi-pi.Cos2x > 0 when -pi/4 < x <pi/4. That is when x is in (npi/2-pi/4 , npi/2+pi/4), n = 0,1,2,.. Therefore sin2x - 90 is increasing in (npi-pi/4 ,npi+ pi/4). When x is in the interval (180n-45 deg ,180n+45 deg ) for n =0,1,2,....Sin2x -90 is decreasing when (sin2x-90)' < 0. Or2cos2x < 0 or cos2x < 0. cos2x < 0 when 2npi+ pi/2 < 2x < 2npi+3pi/2. npi+Pi/4 < x < npi+3pi/4 is same as when x is in (180n+45deg to 135 deg.), n =0,1,2,3....Therefore sin2x-90 is decreasing in (180n+45 deg , 180n+135deg), for n = 0,1,2,3....