Solving a trigonometric equation with cotI am asked to solve the following equation: 6 sin ⁡ ( t ) = cos ⁡ ( 2 t ) − 5 tan ⁡ ( X ) where X is the solution of 8 sin ⁡ ( 2 x ) + cos ⁡ ( 2 x ) = 10 cot ⁡ ( x ) − 2.I have already tried to replace the cotangent by equivalent expressions but I have not been able to quite solve it the way it is actually done in the textbook. Indeed, the way it is presented is by setting z = tan ⁡ ( x ) and solving z 3 + 6 z 2 + 3 z − 10 = 0, thus finding z to be either -5, -2 or 1. I cannot get to this last 3rd-order equation. Any hint would be much appreciated!

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Solving a trigonometric equation with cotI am asked to solve the following equation:   6  sin  ⁡  (  t  )  =            cos      ⁡      (      2      t      )      −      5              tan      ⁡      (      X      )       where X is the solution of   8  sin  ⁡  (  2  x  )  +  cos  ⁡  (  2  x  )  =  10  cot  ⁡  (  x  )  −  2.I have already tried to replace the cotangent by equivalent expressions but I have not been able to quite solve it the way it is actually done in the textbook. Indeed, the way it is presented is by setting   z  =  tan  ⁡  (  x  ) and solving       z    3    +  6      z    2    +  3  z  −  10  =  0, thus finding z to be either -5, -2 or 1. I cannot get to this last 3rd-order equation. Any hint would be much appreciated!

Berattirna · Accepted Answer

Rewrite   8  sin  ⁡  (  2  x  )  +  cos  ⁡  (  2  x  )  =  10  cot  ⁡  (  x  )  −  2 using the formulas that give   sin and   cos in function of   tan of the half angle  sin  ⁡  2  x  =            2      p              1      +              p        2              ;    cos  ⁡  2  x  =            1      −              p        2                    1      +              p        2              ;    p  =  tan  ⁡  x  ;    cot  ⁡  x  =      1    p  the equation becomes            16      p              1      +              p        2              +            1      −              p        2                    1      +              p        2              =      10    p    −  2Which becomes      p    3    +  6      p    2    +  3  p  −  10  =  0factoring we get  (  p  −  1  )  (  p  +  2  )  (  p  +  5  )  =  0  →      p    1    =  1  ;        p    2    =  −  2  ;        p    3    =  −  5This means that we can substitute   tan  ⁡  x  =  1 but not     tan  ⁡  x  =  −  2  ;    tan  ⁡  x  =  −  5 in the first equation  6  sin  ⁡  (  t  )  =            cos      ⁡      (      2      t      )      −      5              tan      ⁡      (      x      )      becomes  6  sin  ⁡  (  t  )  =  cos  ⁡  (  2  t  )  −  5and finally find the solutions for   t.

Annie French · Accepted Answer

HINT: your equation  8  sin  ⁡  (  2  x  )  +  cos  ⁡  (  2  x  )  =  10  cot  ⁡  (  x  )  −  2is equivalent to  2        (    5                      t                    2              −    2        t    −    5    )        (                  t                    2              −    t    −    1    )        (                  t                    2              +    2        t    −    1    )    =  0with  t  =  tan  ⁡  (  x      /    2  )

Solving a trigonometric equation with cot I am asked to solve the following equation: 6 sin(t)=(cos(2t)-5)/(sqrt(tan(X))) where X is the solution of 8sin(2x)+cos(2x)=10cot(x)−2.

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