Sin^2(x)+cos^2(2x)=1

[latex] sin^{2x} +cos^{2} 2x=1 [/latex][latex](1-cos2x) /2 + cos^22x=1[/latex]   домножим на 2 [latex]1-cos2x +2 cos^2 2x=2[/latex] [latex]2 cos^2 2x -cos 2x -1=0 [/latex] заменим  cos2x   на  y [latex]2 y^2 -y -1=0[/latex] D = 9  ;  √D=+-3 y₁=-1/2 ;  cos2x = y₁= -1/2 ;  x =π*n - π/3, n ∈ Z x =pi*n + pi/3, n ∈ Z y₂= 1 ;  cos2x = y₂ =1 ;  x= π*n , n ∈ Z Ответ: x =π*n - π/3, n ∈ Z ; x =π*n + π/3, n ∈ Z ; x= π*n , n ∈ Z