Recall that `sinx + cosx =u` (say)
and `sin x cosx =v` (say) are connected by
`(sinx +cosx)^(2) = sin^(2)x + cos^(2)x+2sin cosx`
`rArr u^(2) = 1+2v`
`rArr v=(u^(2)-1)/(2)`
It follows that any rational integral function of `sinx + cosx`, and `sinx cosx` i.e., `R(sinx + cosx, sinx cosx)`, or in our notation R(u,v) can be transformed to `R(u, (u^(2)-1)/2)`. Thus, to solve an equation of the form `R(u,v)=0`, we form a polynomial equation in u and than look for solutions.
The solution set of
`sinx + cosx -2sqrt(2) sin x cosx=0` is completely described by

Options

A. `x = 2npi + pi/4, 2npi - (5pi)/12, 2npi + (11pi)/12, n in Z`
B. `x= 2npi-pi/4, 2npi+pi/12, 2npi + (7pi)/12, n in Z`
C. `x = 2npi + pi/4, 2npi - pi/12, 2npi - (7pi)/12, n in Z`
D. `x=2npi - pi/4, 2npi + pi/12, 2npi - (7pi)/12, n in Z`

Answer

D