trying to solve cos ⁡ ( x ) − 2 cos ⁡ ( 2 x ) + 2 cos ⁡ ( 2 x ) = 0The system is { cos ⁡ ( x ) − 2 cos ⁡ ( 2 x ) = 2 cos 2 ⁡ ( 2 x ) − 2 cos ⁡ ( 2 x ) ≥ 0 ⟺ cos ⁡ ( 2 x ) ≤ 0 The equation: cos ⁡ ( x ) − 2 ( 2 cos 2 ⁡ ( x ) − 1 ) = 2 ( 2 cos 2 ⁡ ( x ) − 1 ) 2 cos ⁡ ( x ) − 4 cos 2 ⁡ ( x ) + 2 = 2 ( 4 cos 4 ⁡ ( x ) − 4 cos 2 ⁡ ( x ) + 1 ) cos ⁡ ( x ) − 4 cos 2 ⁡ ( x ) + 2 = 8 cos 4 ⁡ ( x ) − 8 cos 2 ⁡ ( x ) + 2 8 cos 4 ⁡ ( x ) − 4 cos 2 ⁡ ( x ) − cos ⁡ ( x ) = 0But what to do next?If I factor out cos ⁡ ( x ), I get a non-factorizable third-degree polynomial in parentheses cos ⁡ ( x ) ( 8 cos 3 ⁡ ( x ) − 4 cos ⁡ ( x ) − 1 ) = 0

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trying to solve       cos    ⁡    (    x    )    −    2    cos    ⁡    (    2    x    )    +      2    cos  ⁡  (  2  x  )  =  0The system is      {                            cos          ⁡          (          x          )          −          2          cos          ⁡          (          2          x          )          =          2                      cos            2                    ⁡          (          2          x          )                                      −                      2                    cos          ⁡          (          2          x          )          ≥          0                    ⟺                    cos          ⁡          (          2          x          )          ≤          0                        The equation:  cos  ⁡  (  x  )  −  2  (  2      cos    2    ⁡  (  x  )  −  1  )  =  2  (  2      cos    2    ⁡  (  x  )  −  1      )    2    cos  ⁡  (  x  )  −  4      cos    2    ⁡  (  x  )  +  2  =  2  (  4      cos    4    ⁡  (  x  )  −  4      cos    2    ⁡  (  x  )  +  1  )  cos  ⁡  (  x  )  −  4      cos    2    ⁡  (  x  )  +  2  =  8      cos    4    ⁡  (  x  )  −  8      cos    2    ⁡  (  x  )  +  2  8      cos    4    ⁡  (  x  )  −  4      cos    2    ⁡  (  x  )  −  cos  ⁡  (  x  )  =  0But what to do next?If I factor out   cos  ⁡  (  x  ), I get a non-factorizable third-degree polynomial in parentheses  cos  ⁡  (  x  )  (  8      cos    3    ⁡  (  x  )  −  4  cos  ⁡  (  x  )  −  1  )  =  0

Braylon Perez · Accepted Answer

HINT:  8      cos    3    ⁡  x  −  4  cos  ⁡  x  −  1  =  4  cos  ⁡  x  (  2      cos    2    ⁡  x  −  1  )    ⟹    2  (  cos  ⁡  3  x  +  cos  ⁡  x  )  =  1    ⟺    4  cos  ⁡  x  cos  ⁡  2  x  =  1If   sin  ⁡  x  =  0  ,  4  cos  ⁡  x  cos  ⁡  2  x  ≠  1    ⟹    sin  ⁡  x  ≠  0  4  cos  ⁡  x  cos  ⁡  2  x  =  1    ⟺    sin  ⁡  x  =  2  (  2  sin  ⁡  x  cos  ⁡  x  )  cos  ⁡  2  x  =  2  sin  ⁡  2  x  cos  ⁡  2  x  =  sin  ⁡  4  x  sin  ⁡  4  x  =  sin  ⁡  x    ⟹    4  x  =  n  π  +  (  −  1      )    n    xwhere n is any integerAlso,   cos  ⁡  2  x  ≤  0    ⟹    2  m  π  +            π      2        ≤  2  x  ≤  2  m  π  +                    3        π            2

trying to solve <msqrt> cos ⁡ ( x ) −

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