How would I solve this trigonometric equation ln ⁡ [ ( sin ⁡ ( x ) + cos ⁡ ( x ) ) 1 + sin ⁡ ( 2 x ) ] = 2?I know that 1 + sin ⁡ ( 2 x ) = ( sin ⁡ ( x ) + cos ⁡ ( x ) ) 2 , but I cant understand which property of logarithms to apply to get the solution here.

Question

How would I solve this trigonometric equation   ln  ⁡  [  (  sin  ⁡  (  x  )  +  cos  ⁡  (  x  )      )          1      +      sin      ⁡      (      2      x      )        ]  =  2?I know that   1  +  sin  ⁡  (  2  x  )  =  (  sin  ⁡  (  x  )  +  cos  ⁡  (  x  )      )    2  , but I cant understand which property of logarithms to apply to get the solution here.

reinmelk3iu · Accepted Answer

The equation can be written as:      t                  t        2              =      e    2  where   t  =  sin  ⁡  x  +  cos  ⁡  x. The function   f  :  (  0  ,      2    ]  →      R    ,  f  (  t  )  =      t                  t        2            , has one local minimum at   t  =      1          e      , so:  f  (  t  )  ≤  max  {      lim          x      →              0        +              f  (  x  )  ,  f  (      2    )  }  =  f  (      2    )  =  2  &amp;lt;      e    2  because       lim          x      →              0        +              f  (  x  )  =  1. The equation does not have any solution.

How would I solve this trigonometric equation ln[(sin(x)+cos(x))^(1+sin(2x))]=2?

Answered question

Answer & Explanation

New Questions in Trigonometry