Question

1-4sin^2t*cos^2t/(sint+cost)^2+2sint*cost=1

Answer 1

Во  всей вероятности 
(1- 4sin²t·cos²t)/(sint+cost)² + 2sint·cost) =1
 (1- sin2t)²/(1+2sint·cost) +sin2t = 1
  1- sin2t + sin2t + (sin2t)² = 1+sin2t    ; 
  (sin2t)² - sin2t =0
  sin2t·(sin2t -1) = 0
   a)   sin2t =0    ⇒  2t = πk  ⇒                 t= π/2 ·k    ;  k∈Z
   b)   sin2t =1    ⇒  2t = 2πm + π/2   ⇒    t=πm +π/4 ; m∈Z