1.

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 of sinx + cosx -2sqrt(2) =0 is completely described by

Answer»

`x = 2NPI + pi/4, 2npi - (5PI)/12, 2npi + (11pi)/12, n in Z`
`x= 2npi-pi/4, 2npi+pi/12, 2npi + (7pi)/12, n in Z`
`x = 2npi+ pi/4, 2npi - pi/12, 2npi - (7pi)/12, n in Z`
`x=2npi - pi/4, 2npi + pi/12, 2npi - (7pi)/12, n in Z`

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