InterviewSolution
Saved Bookmarks
| 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` |
|