Correct Answer - C
Given, `sin^(2) 2 theta + cos ^(4) 2 theta = (3)/(4)`
`rArr (1-cos^(2) 2 theta) + (cos^(4) 2 theta = (3)/(4) " "(because sin^(2) x = 1 - cos^(2) x)`
`rArr 4 cos^(4) 2 theta - 4 cos^(2) theta +1= 0`
`rArr ( 2 cos ^(2) 2theta - 1)^(2) = 0`
`rArr 2 cos 2 theta -1 = 0`
`rArr cos^(2) 2 theta = (1)/(2)`
`rArr cos 2 theta = pm (1)/(sqrt(2))`
If `theta in (0,(pi)/(2))`, then `2 theta in (0,pi)`
`thereforecos 2 theta = (pi)/(sqrt(2))`
`rArr 2 theta = (pi)/(4),(3pi)/(2), [ because cos((3pi)/(4))= cos (pi-(pi)/(4)) = - cos .(pi)/(4) = -(1)/(sqrt(2))]`
`rArr theta = (pi)/(8),(3pi)/(8)`
Sum of value of `theta= (pi)/(8) + (3pi)/(8) = (pi)/(2)`
