Finding the range of 5 cos ⁡ x − 2 sin 2 ⁡ x + 4 sin ⁡ x − 3 6 | cos ⁡ x | + 1

Question

Finding the range of             5      cos      ⁡      x      −      2              sin                  2                    ⁡      x      +      4      sin      ⁡      x      −      3              6              |            cos      ⁡      x              |            +      1

cokeman206 · Accepted Answer

If   x  ∈  [  −  π      /    2  ,  π      /    2  ] then       |    cos  ⁡  x      |    =  cos  ⁡  x so            5      cos      ⁡      x      −      2              sin                  2                    ⁡      x      +      4      sin      ⁡      x      −      3              6              |            cos      ⁡      x              |            +      1        =                    5        cos        ⁡        x        +        5                  /                6                    6        cos        ⁡        x        +        1              +                    −        2                  sin          2                ⁡        x        +        4        sin        ⁡        x        −        3                  5          6                            6        cos        ⁡        x        +        1                =  5      /    6  +                    −        2                  sin          2                ⁡        x        +        4        sin        ⁡        x        −        3                  5          6                            6        cos        ⁡        x        +        1              The   −  2      sin    2    ⁡  x term is always negative, so let   x  =  −  π      /    2 then   sin  ⁡  x  =  −  1,   cos  ⁡  x  =  0, and so by inspection it is clearly the minimum and the value is −9.You can apply the same argument to the other range   x  ∈  [  π      /    2  ,  3  π      /    2  ] and the same minimum is found.

Finding the range of (5 cos x−2 sin^2 x+4 sin x−3)/(6|cos x|+1)

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