Solve trigonometric inequality 2 cos 2 ⁡ x + cos ⁡ ( 2 x ) ≥ 1My attempt: 2 cos 2 ⁡ x + cos 2 ⁡ x − sin 2 ⁡ x ≥ sin 2 ⁡ x + cos 2 ⁡ x 2 cos 2 ⁡ x − 2 sin 2 ⁡ x ≥ 0 2 ( cos 2 ⁡ x − sin 2 ⁡ x ) ≥ 0 2 cos ⁡ ( 2 x ) ≥ 0 cos ⁡ ( 2 x ) ≥ 0What`s next?

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Solve trigonometric inequality   2      cos    2    ⁡  x  +  cos  ⁡  (  2  x  )  ≥  1My attempt:  2      cos    2    ⁡  x  +      cos    2    ⁡  x  −      sin    2    ⁡  x  ≥      sin    2    ⁡  x  +      cos    2    ⁡  x  2      cos    2    ⁡  x  −  2      sin    2    ⁡  x  ≥  0  2  (      cos    2    ⁡  x  −      sin    2    ⁡  x  )  ≥  0  2  cos  ⁡  (  2  x  )  ≥  0  cos  ⁡  (  2  x  )  ≥  0What`s next?

postotnojeyf · Accepted Answer

You ended up with                     (1)                    cos        ⁡        2        x        ≥        0            This is possible if and only if 2x is in either first or the fourth quadrant. In particular,(1) is true if   2  x  ∈      [    0    ,          π      2        ]    ∪    [                  3        π            2        ,    2    π    ]  To generalize the blue colored domain, just add   2  n  π  ,  n  ∈      Z   to get the following:  2  x  ∈  [  2  n  π  ,  2  n  π  +      π    2    ]  ∪  [  2  n  π  +            3      π        2    ,  2  n  π  +  2  π  ]  ,  n  ∈      Z  It follows that   x  ∈  [  n  π  ,  n  π  +      π    4    ]  ∪  [  n  π  +            3      π        4    ,  n  π  +  π  ]  ,  n  ∈      Z

Solve trigonometric inequality 2 cos^2 x+cos(2x) >= 1

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