Correct Answer - D
`cos 2theta = (sqrt(2)+1)(cos. theta-(1)/(sqrt(2)))`
`rArr (2 cos^(2)theta-1)=(sqrt(2)+1)/(sqrt(2))(sqrt(2)cos theta -1)`
`rArr (sqrt(2)cos theta -1)(sqrt(2)cos theta +1)=(sqrt(2)+1)/(sqrt(2))(sqrt(2)cos theta -1)`
`rArr cos. theta =(1)/(sqrt(2))` or `cos. theta =(1)/(2)`
`rArr theta = 2n pi pm (pi)/(4)` or `theta = 2n pi pm(pi)/(3), n in Z`