Solve the equation 1 sin 2 k ⁡ ( x ) + 1 cos 2 k ⁡ ( x ) = 8 where k is an integer and x is a real number.

Question

Solve the equation       1                  sin                  2          k                    ⁡      (      x      )        +      1                  cos                  2          k                    ⁡      (      x      )        =  8 where k is an integer and x is a real number.

poquetahr · Accepted Answer

We have  8  =            1                        sin                      2            k                          ⁡        (        x        )              +            1                        cos                      2            k                          ⁡        (        x        )              ≥            2              |                  sin          k                ⁡        (        x        )                  cos          k                ⁡        (        x        )        |              =                    2                  k          +          1                            |                  sin          k                ⁡        (        2        x        )        |              ≥      2          k      +      1      Hence, we have   k  +  1  ≤  3    ⟹    k  ≤  2If k=0, we have             1                        sin                      2            k                          ⁡        (        x        )              +            1                        cos                      2            k                          ⁡        (        x        )              =  2If k&amp;lt;0, we have      sin          −      2      k        ⁡  (  x  )  +      cos          −      2      k        ⁡  (  x  )  ≤      sin    2    ⁡  (  x  )  +      cos    2    ⁡  (  x  )  =  1Hence, the only options are k=1 and k=2.If k=2, we see that      sin          −      2      k        ⁡  (  x  )  +      cos          −      2      k        ⁡  (  x  )  ≤      sin    2    ⁡  (  x  )  +      cos    2    ⁡  (  x  )  =  1This implies that  2  x  =  n  π  +            π      2          ⟹    x  =                    n        π            2        +            π      4      When k=1, we have that            1                        sin          2                ⁡        (        x        )              +            1                        cos          2                ⁡        (        x        )              =            1                        sin          2                ⁡        (        x        )                  cos          2                ⁡        (        x        )              =            4                        sin          2                ⁡        (        2        x        )              =  8Hence,      sin    2    ⁡  (  2  x  )  =            1      2          ⟹    2  x  =  n  π  ±            π      4          ⟹    x  =                    n        π            2        ±            π      8

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