How to solve equations like $alpha sin x -betasin 2x +gamma=0 $

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












Can I solve this equation without Newton-Raphson method?
I have $alpha=47.02$ $beta=112.5$ and $gamma=50$.
When I have to use Newton-Rapson to solve trigonometric equations ?
I will greatly appreciate your answers.







share|cite|improve this question















  • 1




    Are you asking for a type of closed form for x? Are you asking for different numerical methods?
    – Mason
    Jul 19 at 21:36














up vote
1
down vote

favorite












Can I solve this equation without Newton-Raphson method?
I have $alpha=47.02$ $beta=112.5$ and $gamma=50$.
When I have to use Newton-Rapson to solve trigonometric equations ?
I will greatly appreciate your answers.







share|cite|improve this question















  • 1




    Are you asking for a type of closed form for x? Are you asking for different numerical methods?
    – Mason
    Jul 19 at 21:36












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Can I solve this equation without Newton-Raphson method?
I have $alpha=47.02$ $beta=112.5$ and $gamma=50$.
When I have to use Newton-Rapson to solve trigonometric equations ?
I will greatly appreciate your answers.







share|cite|improve this question











Can I solve this equation without Newton-Raphson method?
I have $alpha=47.02$ $beta=112.5$ and $gamma=50$.
When I have to use Newton-Rapson to solve trigonometric equations ?
I will greatly appreciate your answers.









share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 19 at 21:33









Mauricio J. S.

61




61







  • 1




    Are you asking for a type of closed form for x? Are you asking for different numerical methods?
    – Mason
    Jul 19 at 21:36












  • 1




    Are you asking for a type of closed form for x? Are you asking for different numerical methods?
    – Mason
    Jul 19 at 21:36







1




1




Are you asking for a type of closed form for x? Are you asking for different numerical methods?
– Mason
Jul 19 at 21:36




Are you asking for a type of closed form for x? Are you asking for different numerical methods?
– Mason
Jul 19 at 21:36










2 Answers
2






active

oldest

votes

















up vote
3
down vote













Hint:
Substitute
$$sin(x)=frac2t1+t^2$$



$$cos(x)=frac1-t^21+t^2$$
the so called Weierstrass substitution.






share|cite|improve this answer



















  • 2




    This is erroneously called the Weierstrass substitution. James Stewart's widely used calculus textbooks say that Weierstrass pointed it out. Stewart claimed to have learned that elsewhere but said he couldn't remember where. That's the blind leading the blind. I'd call it the tangent half-angle substitution, since $t = tandfrac x 2. qquad$
    – Michael Hardy
    Jul 19 at 22:06











  • @MichaelHardy, Indeed! I would say that your comment might only be improved by a link to wiki en.wikipedia.org/wiki/Tangent_half-angle_substitution which confirms this.
    – Mason
    Jul 19 at 22:09











  • Ok, let's say tan half angle substitution.
    – Dr. Sonnhard Graubner
    Jul 19 at 22:25

















up vote
1
down vote













Alt. hint:   write it as $, 2 beta sin x cos x = alpha sin x + gamma,$, and square both sides. With $,s = sin x,$ the equation then becomes a depressed quartic: $;4 beta^2 s^2(1-s^2) = (alpha s + gamma)^2,$.






share|cite|improve this answer





















    Your Answer




    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );








     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857066%2fhow-to-solve-equations-like-alpha-sin-x-beta-sin-2x-gamma-0%23new-answer', 'question_page');

    );

    Post as a guest






























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote













    Hint:
    Substitute
    $$sin(x)=frac2t1+t^2$$



    $$cos(x)=frac1-t^21+t^2$$
    the so called Weierstrass substitution.






    share|cite|improve this answer



















    • 2




      This is erroneously called the Weierstrass substitution. James Stewart's widely used calculus textbooks say that Weierstrass pointed it out. Stewart claimed to have learned that elsewhere but said he couldn't remember where. That's the blind leading the blind. I'd call it the tangent half-angle substitution, since $t = tandfrac x 2. qquad$
      – Michael Hardy
      Jul 19 at 22:06











    • @MichaelHardy, Indeed! I would say that your comment might only be improved by a link to wiki en.wikipedia.org/wiki/Tangent_half-angle_substitution which confirms this.
      – Mason
      Jul 19 at 22:09











    • Ok, let's say tan half angle substitution.
      – Dr. Sonnhard Graubner
      Jul 19 at 22:25














    up vote
    3
    down vote













    Hint:
    Substitute
    $$sin(x)=frac2t1+t^2$$



    $$cos(x)=frac1-t^21+t^2$$
    the so called Weierstrass substitution.






    share|cite|improve this answer



















    • 2




      This is erroneously called the Weierstrass substitution. James Stewart's widely used calculus textbooks say that Weierstrass pointed it out. Stewart claimed to have learned that elsewhere but said he couldn't remember where. That's the blind leading the blind. I'd call it the tangent half-angle substitution, since $t = tandfrac x 2. qquad$
      – Michael Hardy
      Jul 19 at 22:06











    • @MichaelHardy, Indeed! I would say that your comment might only be improved by a link to wiki en.wikipedia.org/wiki/Tangent_half-angle_substitution which confirms this.
      – Mason
      Jul 19 at 22:09











    • Ok, let's say tan half angle substitution.
      – Dr. Sonnhard Graubner
      Jul 19 at 22:25












    up vote
    3
    down vote










    up vote
    3
    down vote









    Hint:
    Substitute
    $$sin(x)=frac2t1+t^2$$



    $$cos(x)=frac1-t^21+t^2$$
    the so called Weierstrass substitution.






    share|cite|improve this answer















    Hint:
    Substitute
    $$sin(x)=frac2t1+t^2$$



    $$cos(x)=frac1-t^21+t^2$$
    the so called Weierstrass substitution.







    share|cite|improve this answer















    share|cite|improve this answer



    share|cite|improve this answer








    edited Jul 19 at 21:58


























    answered Jul 19 at 21:46









    Dr. Sonnhard Graubner

    66.8k32659




    66.8k32659







    • 2




      This is erroneously called the Weierstrass substitution. James Stewart's widely used calculus textbooks say that Weierstrass pointed it out. Stewart claimed to have learned that elsewhere but said he couldn't remember where. That's the blind leading the blind. I'd call it the tangent half-angle substitution, since $t = tandfrac x 2. qquad$
      – Michael Hardy
      Jul 19 at 22:06











    • @MichaelHardy, Indeed! I would say that your comment might only be improved by a link to wiki en.wikipedia.org/wiki/Tangent_half-angle_substitution which confirms this.
      – Mason
      Jul 19 at 22:09











    • Ok, let's say tan half angle substitution.
      – Dr. Sonnhard Graubner
      Jul 19 at 22:25












    • 2




      This is erroneously called the Weierstrass substitution. James Stewart's widely used calculus textbooks say that Weierstrass pointed it out. Stewart claimed to have learned that elsewhere but said he couldn't remember where. That's the blind leading the blind. I'd call it the tangent half-angle substitution, since $t = tandfrac x 2. qquad$
      – Michael Hardy
      Jul 19 at 22:06











    • @MichaelHardy, Indeed! I would say that your comment might only be improved by a link to wiki en.wikipedia.org/wiki/Tangent_half-angle_substitution which confirms this.
      – Mason
      Jul 19 at 22:09











    • Ok, let's say tan half angle substitution.
      – Dr. Sonnhard Graubner
      Jul 19 at 22:25







    2




    2




    This is erroneously called the Weierstrass substitution. James Stewart's widely used calculus textbooks say that Weierstrass pointed it out. Stewart claimed to have learned that elsewhere but said he couldn't remember where. That's the blind leading the blind. I'd call it the tangent half-angle substitution, since $t = tandfrac x 2. qquad$
    – Michael Hardy
    Jul 19 at 22:06





    This is erroneously called the Weierstrass substitution. James Stewart's widely used calculus textbooks say that Weierstrass pointed it out. Stewart claimed to have learned that elsewhere but said he couldn't remember where. That's the blind leading the blind. I'd call it the tangent half-angle substitution, since $t = tandfrac x 2. qquad$
    – Michael Hardy
    Jul 19 at 22:06













    @MichaelHardy, Indeed! I would say that your comment might only be improved by a link to wiki en.wikipedia.org/wiki/Tangent_half-angle_substitution which confirms this.
    – Mason
    Jul 19 at 22:09





    @MichaelHardy, Indeed! I would say that your comment might only be improved by a link to wiki en.wikipedia.org/wiki/Tangent_half-angle_substitution which confirms this.
    – Mason
    Jul 19 at 22:09













    Ok, let's say tan half angle substitution.
    – Dr. Sonnhard Graubner
    Jul 19 at 22:25




    Ok, let's say tan half angle substitution.
    – Dr. Sonnhard Graubner
    Jul 19 at 22:25










    up vote
    1
    down vote













    Alt. hint:   write it as $, 2 beta sin x cos x = alpha sin x + gamma,$, and square both sides. With $,s = sin x,$ the equation then becomes a depressed quartic: $;4 beta^2 s^2(1-s^2) = (alpha s + gamma)^2,$.






    share|cite|improve this answer

























      up vote
      1
      down vote













      Alt. hint:   write it as $, 2 beta sin x cos x = alpha sin x + gamma,$, and square both sides. With $,s = sin x,$ the equation then becomes a depressed quartic: $;4 beta^2 s^2(1-s^2) = (alpha s + gamma)^2,$.






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        Alt. hint:   write it as $, 2 beta sin x cos x = alpha sin x + gamma,$, and square both sides. With $,s = sin x,$ the equation then becomes a depressed quartic: $;4 beta^2 s^2(1-s^2) = (alpha s + gamma)^2,$.






        share|cite|improve this answer













        Alt. hint:   write it as $, 2 beta sin x cos x = alpha sin x + gamma,$, and square both sides. With $,s = sin x,$ the equation then becomes a depressed quartic: $;4 beta^2 s^2(1-s^2) = (alpha s + gamma)^2,$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 20 at 0:15









        dxiv

        54.2k64797




        54.2k64797






















             

            draft saved


            draft discarded


























             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857066%2fhow-to-solve-equations-like-alpha-sin-x-beta-sin-2x-gamma-0%23new-answer', 'question_page');

            );

            Post as a guest













































































            Comments

            Popular posts from this blog

            What is the equation of a 3D cone with generalised tilt?

            Color the edges and diagonals of a regular polygon

            Relationship between determinant of matrix and determinant of adjoint?