the following equality is given: 2 2 sin ⁡ x + 2 cos ⁡ x = − sin ⁡ 2 x

Question

the following equality is given:  2      2    sin  ⁡      x  +      2    cos  ⁡      x    =      −    sin    ⁡    2    x

Elias Flores · Accepted Answer

Since       (    2          2              )      2        +    (          2              )      2        =      10  , we have                    2                  2                sin        ⁡        x        +                  2                cos        ⁡        x                            =                  10                          (                                    2                              2                                                    10                                sin          ⁡          x          +                                    2                                      10                                cos          ⁡          x          )                                                  =                  10                          (                      2                          5                                sin          ⁡          x          +                      1                          5                                cos          ⁡          x          )                                                  =                  10                (        cos        ⁡        θ        sin        ⁡        x        +        sin        ⁡        θ        cos        ⁡        x        )                                          =                  10                sin        ⁡        (        x        +        θ        )            where  cos  ⁡  θ  =      2          5        ,    sin  ⁡  θ  =      1          5          ⇒    tan  ⁡  θ  =      1    2      ⇒    θ  =  arctan  ⁡  (  1      /    2  )Thus, for   k  ∈      Z                            2                  2                sin        ⁡        x        +                  2                cos        ⁡        x        ≥        0                                                  ⟺                          10                sin        ⁡        (        x        +        arctan        ⁡        (        1                  /                2        )        )        ≥        0                                                  ⟺                0        +        2        k        π        ≤        x        +        arctan        ⁡        (        1                  /                2        )        ≤        π        +        2        k        π                                                  ⟺                −        arctan        ⁡        (        1                  /                2        )        +        2        k        π        ≤        x        ≤        π        −        arctan        ⁡        (        1                  /                2        )        +        2        k        π            We can see that this is the same as  2  arctan  ⁡  (  2  −      5    )  +  2  k  π  ≤  x  ≤  2  arctan  ⁡  (  2  +      5    )  +  2  k  πbecause                          2        arctan        ⁡        (        2        −                  5                )        +        arctan        ⁡        (        1                  /                2        )                                          =        arctan        ⁡        (        2        −                  5                )        +        arctan        ⁡        (        2        −                  5                )        +        arctan        ⁡        (        1                  /                2        )                                          =        arctan        ⁡        (        2        −                  5                )        +        arctan        ⁡                  (                                    2              −                              5                            +              (              1                              /                            2              )                                      1              −              (              2              −                              5                            )                              /                            2                                )                                                  =        arctan        ⁡        (        2        −                  5                )        +        arctan        ⁡        (                  5                −        2        )                                          =        0            and                          2        arctan        ⁡        (        2        +                  5                )        +        arctan        ⁡        (        1                  /                2        )                                          =        arctan        ⁡        (        2        +                  5                )        +        arctan        ⁡        (        2        +                  5                )        +        arctan        ⁡        (        1                  /                2        )                                          =        arctan        ⁡        (        2        +                  5                )        +        π        +        arctan        ⁡                  (                                    2              +                              5                            +              (              1                              /                            2              )                                      1              −              (              2              +                              5                            )                              /                            2                                )                                                  =        arctan        ⁡        (        2        +                  5                )        +        π        +        arctan        ⁡        (        −        2        −                  5                )                                          =        π

Blericker74 · Accepted Answer

Clearly,   sin  ⁡  x  cos  ⁡  x  ≠  0We first need   sin  ⁡  x  cos  ⁡  x  &amp;lt;  0    ⟹    tan  ⁡  x  &amp;lt;  0  2      2    sin  ⁡  x  +      2    cos  ⁡  x  =      −    2    sin    ⁡    x    cos    ⁡    x  Divide both sides by       −    2    sin    ⁡    x    cos    ⁡    x    , to get  2      −    tan    ⁡    x    +      −    cot    ⁡    x    =  1Let       −    tan    ⁡    x    =  y  &amp;gt;  0  ,  tan  ⁡  x  =  −      y    2    ,      −    cot    ⁡    x    =            1      y        &amp;gt;  0  2  y  +            1      y        =  1    ⟺    2      y    2    −  y  +  1  =  0    ⟹    y  =                    1        ±                  1          −          8                    4      But   y  &amp;gt;  0

the following equality is given: 2 <msqrt> 2 </msqrt> sin ⁡ <mrow

Answered question

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