Methodology to solve trigonometric equations such as 4 x + tan ⁡ ( x ) = πI have to solve the following trigonometric equation: 4 ⋅ arctan ⁡ ( ω 180 ) + ω 180 = πwhich by setting arctan ⁡ ( ω 180 ) = x becomes: 4 x + tan ⁡ ( x ) − π = 0In order to solve the latter equation in terms of x, I tried to replace the term of tan ⁡ ( x ) by the Taylor series approximation of sin ⁡ ( x ) , cos ⁡ ( x ) using only the first two terms but the derived results were not correct. Anyway, I used MATLAB to solve it and it did solved it. The correct result is: x = 0.61048 ⇒ ω 180 = 0.7 rad/secWhat I would like to ask is in which ways can I solve this equation by hand ? I don't want a complete solution as an answer but rather the methodologies I could study and then apply in such trigonometric equations.

Question

Methodology to solve trigonometric equations such as   4  x  +  tan  ⁡  (  x  )  =  πI have to solve the following trigonometric equation:  4  ⋅  arctan  ⁡  (      ω          180        )     +         ω          180           =     πwhich by setting   arctan  ⁡  (      ω          180        )  =  x becomes:  4  x     +     tan  ⁡  (  x  )     −     π     =     0In order to solve the latter equation in terms of x, I tried to replace the term of    tan  ⁡  (  x  ) by the Taylor series approximation of   sin  ⁡  (  x  )  ,  cos  ⁡  (  x  ) using only the first two terms but the derived results were not correct. Anyway, I used MATLAB to solve it and it did solved it. The correct result is:  x  =  0.61048  ⇒      ω          180        =  0.7  rad/secWhat I would like to ask is in which ways can I solve this equation by hand ? I don&#039;t want a complete solution as an answer but rather the methodologies I could study and then apply in such trigonometric equations.

mignonechatte00f · Accepted Answer

Since it is transcendental, there is no explicit expression for the zero of function  f  (  x  )  =  4  x  +  tan  ⁡  (  x  )  −  πand numerical methods (or approximations) are required.Being very lazy, expand   f  (  x  ) as a series around   x  =  0 and obtain  f  (  x  )  =  −  π  +  5  x  +            x      3        3    +            2              x        5              15    +            17              x        7              315    +            62              x        9              2835    +            1382              x                  11                      155925    +  O      (          x              13              )  Now, using series reversion,  x  =  t  −            t      3        15    −            t      5        75    −            t      7        7875    +            67              t        9              70875    +            5017              t                  11                      19490625    +  O      (          t              13              )  where   t  =            f      (      x      )      +      π        5  . Since we desire   f  (  x  )  =  0 then   t  =      π    5  . Replacing, the estimate is then  x  ∼      0.610487    235while the exact solution given by Newton method is   0.610487119Just for the fun of doing it almost by hand, all of the above was done using my   50 years old non-programmable calculator.

Methodology to solve trigonometric equations such as 4x+tan(x)=pi

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