Cos2x(2cos²2x -1) =1/4 ;
2cos³2x -cos2x -1/4 =0;
t =cos2x;
8t³ -4t -1 =0;
8t³ +4t² - 4t² - 2t - 2t -1 =0;
4t²(2t+1) -2t(2t+1) -(2t+1) =0;
(2t +1)(4t² -2t -1)=0;
a) (2t +1) =0 ;
t₁=-1/2;
cos2x = -1/2 ;
2x =(+/-)π/3 +2π*k , k∈ Z .
x₁=(+/-)π/6+π*k , k∈ Z .
b) 4t² -2t -1 =0;
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t₂=(1+√5)/4;
cos2x =(1+√5)/4;
2x₂ = (+/-)arccos(1+√5)/4) +2π*k ;
x₂= (+/-)arccos((1+√5)/4 )/2 +π*k , k∈ Z .
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t₃=(1-√5)/4;
2x₃= (+/-)arccos(1-√5)/4) +2π*k
x₃= (+/-)arccos((1- √5)/4 )/2 +π*k , k∈ Z .