The solutions of the equation `1+(sin x - cos x)"sin" pi/4=2 "cos"^(2) (5 x)/2` is/are

Options

A. `x=(n pi)/3+pi/8, n in Z`
B. `x=(npi)/2+(5pi)/16, n in Z`
C. `x=(npi)/3+pi/4, n in Z`
D. `x=(npi)/2+(7pi)/8, n in Z`

Answer

A, B

Explanation

`1+(sin x - cos x) "sin" pi/4=2 "cos"^(2) (5pi)/2` ...(i)
`rArr 1+(sin x- cos x) 1/sqrt(2) =1 + cos 5x`
`rArr cos 5x+ cos (x+pi/4)=0`
`rArr 2 cos(3x+pi/8) cos (2x-pi/8)=0`
`rArr cos (3x+pi/8) =0, or cos (2x-pi/8)=0` ...(ii)
`3x+pi/8=n pi +pi/2 or 2x- pi/8 = n pi +pi/2` ...(iii)
or `x=(n pi)/3+pi/8 or x=(n pi)/2+(5 pi)/16, n in Z`