It is given that the acute ang

I think the answer should be √7 / (10√2) .
25cos2x - 20cosx + 2 = 0
25(2cos²x - 1) - 20cosx + 2 = 0
50cos²x - 20cosx - 23 = 0
50(1 - 2sin² x/2)² - 20(1 - 2sin² x/2) - 23 = 0
200sin⁴x/2 - 200sin² x/2 + 50 - 20 + 40sin² x/2 - 23 = 0
200sin⁴x/2 - 160sin² x/2 + 7 = 0
Product of roots = sin² A/2 sin² B/2 = 7 / 200 sin A/2 sin B/2
= √7 / (10√2) or - √7 / (10√2) (rejected since A/2 , B/2 are acute angles)