Solving $sin^2 x +1=2x$












5














How do I solve this equation?



$$sin^2 x +1=2x$$



I have no idea how to attack the problem.



Thanks!






calculus trigonometry







share|cite|improve this question
























  • Since $x$ is both inside and outside of the sine, this is a transcendental equation. You'll only get approximate answers, say, using numerical methods.
    – Blue
    Nov 21 '18 at 17:01






  • 1




    Use numerical methods or wolframalpha.com
    – Vasya
    Nov 21 '18 at 17:01








  • 1




    This equation does not have a nice algebraic solution. You can use numerical methods to find an approximation. If you draw a graph of $sin^x +1 -2x$ you can see about where it crosses the $x$ axis.
    – Ethan Bolker
    Nov 21 '18 at 17:02










  • @Blue Well, sometimes these equations do have nice solutions e.g. $sin x=x$. It doesn't seem to be the case here
    – A.Γ.
    Nov 21 '18 at 17:04










  • @AweKumarJha did you mean Aryabhata?
    – Mohammad Zuhair Khan
    Nov 21 '18 at 17:05
















5














How do I solve this equation?



$$sin^2 x +1=2x$$



I have no idea how to attack the problem.



Thanks!






calculus trigonometry







share|cite|improve this question
























  • Since $x$ is both inside and outside of the sine, this is a transcendental equation. You'll only get approximate answers, say, using numerical methods.
    – Blue
    Nov 21 '18 at 17:01






  • 1




    Use numerical methods or wolframalpha.com
    – Vasya
    Nov 21 '18 at 17:01








  • 1




    This equation does not have a nice algebraic solution. You can use numerical methods to find an approximation. If you draw a graph of $sin^x +1 -2x$ you can see about where it crosses the $x$ axis.
    – Ethan Bolker
    Nov 21 '18 at 17:02










  • @Blue Well, sometimes these equations do have nice solutions e.g. $sin x=x$. It doesn't seem to be the case here
    – A.Γ.
    Nov 21 '18 at 17:04










  • @AweKumarJha did you mean Aryabhata?
    – Mohammad Zuhair Khan
    Nov 21 '18 at 17:05














5












5








5


2





How do I solve this equation?



$$sin^2 x +1=2x$$



I have no idea how to attack the problem.



Thanks!






calculus trigonometry







share|cite|improve this question















How do I solve this equation?



$$sin^2 x +1=2x$$



I have no idea how to attack the problem.



Thanks!






calculus trigonometry




calculus trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 '18 at 16:59









Blue

47.7k870151




47.7k870151










asked Nov 21 '18 at 16:51









L. G.

394




394












  • Since $x$ is both inside and outside of the sine, this is a transcendental equation. You'll only get approximate answers, say, using numerical methods.
    – Blue
    Nov 21 '18 at 17:01






  • 1




    Use numerical methods or wolframalpha.com
    – Vasya
    Nov 21 '18 at 17:01








  • 1




    This equation does not have a nice algebraic solution. You can use numerical methods to find an approximation. If you draw a graph of $sin^x +1 -2x$ you can see about where it crosses the $x$ axis.
    – Ethan Bolker
    Nov 21 '18 at 17:02










  • @Blue Well, sometimes these equations do have nice solutions e.g. $sin x=x$. It doesn't seem to be the case here
    – A.Γ.
    Nov 21 '18 at 17:04










  • @AweKumarJha did you mean Aryabhata?
    – Mohammad Zuhair Khan
    Nov 21 '18 at 17:05


















  • Since $x$ is both inside and outside of the sine, this is a transcendental equation. You'll only get approximate answers, say, using numerical methods.
    – Blue
    Nov 21 '18 at 17:01






  • 1




    Use numerical methods or wolframalpha.com
    – Vasya
    Nov 21 '18 at 17:01








  • 1




    This equation does not have a nice algebraic solution. You can use numerical methods to find an approximation. If you draw a graph of $sin^x +1 -2x$ you can see about where it crosses the $x$ axis.
    – Ethan Bolker
    Nov 21 '18 at 17:02










  • @Blue Well, sometimes these equations do have nice solutions e.g. $sin x=x$. It doesn't seem to be the case here
    – A.Γ.
    Nov 21 '18 at 17:04










  • @AweKumarJha did you mean Aryabhata?
    – Mohammad Zuhair Khan
    Nov 21 '18 at 17:05
















Since $x$ is both inside and outside of the sine, this is a transcendental equation. You'll only get approximate answers, say, using numerical methods.
– Blue
Nov 21 '18 at 17:01




Since $x$ is both inside and outside of the sine, this is a transcendental equation. You'll only get approximate answers, say, using numerical methods.
– Blue
Nov 21 '18 at 17:01




1




1




Use numerical methods or wolframalpha.com
– Vasya
Nov 21 '18 at 17:01






Use numerical methods or wolframalpha.com
– Vasya
Nov 21 '18 at 17:01






1




1




This equation does not have a nice algebraic solution. You can use numerical methods to find an approximation. If you draw a graph of $sin^x +1 -2x$ you can see about where it crosses the $x$ axis.
– Ethan Bolker
Nov 21 '18 at 17:02




This equation does not have a nice algebraic solution. You can use numerical methods to find an approximation. If you draw a graph of $sin^x +1 -2x$ you can see about where it crosses the $x$ axis.
– Ethan Bolker
Nov 21 '18 at 17:02












@Blue Well, sometimes these equations do have nice solutions e.g. $sin x=x$. It doesn't seem to be the case here
– A.Γ.
Nov 21 '18 at 17:04




@Blue Well, sometimes these equations do have nice solutions e.g. $sin x=x$. It doesn't seem to be the case here
– A.Γ.
Nov 21 '18 at 17:04












@AweKumarJha did you mean Aryabhata?
– Mohammad Zuhair Khan
Nov 21 '18 at 17:05




@AweKumarJha did you mean Aryabhata?
– Mohammad Zuhair Khan
Nov 21 '18 at 17:05










4 Answers
4






active

oldest

votes


















5














$$1le 1+sin^2(x)le 2implies frac 12le xle 1.$$
the function
$$f:xmapsto sin^2(x)+1-2x$$
is continuous at $[frac 12,1],$



$$f(frac 12)>0; ; f(1)<0,$$
and



$$f'(x)=sin(2x)-2<0.$$



By IVT,
there is a unique solution $alpha$ in $]frac 12,1[$.



$$alpha=lim_{nto+infty}u_n$$



with
$$u_0=1$$
and
$$u_{n+1}=u_n-frac{f(u_n)}{f'(u_n)}$$



This numerical method is known as Newton-Raphson.






share|cite|improve this answer































    2














    Once it is proved that $f(x)=sin^2(x)-2x+1$ has a unique real zero in the interval $[0,pi/4]$, its numerical determination is simple since $f(x)$ is a positive and convex function on $(0,pi/4)$ (due to $f''(x)>0$), hence by applying Newton's method with starting point $x=0$ we get



    $$ rho approx 0.714836$$
    in just four steps.






    share|cite|improve this answer





























      1














      Just for the fun of the approximation.



      Using the double angle formula Rewrite the equation as $$cos(2x)+4x=3$$ Now, let $t=2x$ to make the equation
      $$cos(t)+2t=3$$
      Now, using the approximation
      $$cos(t) simeqfrac{pi ^2-4t^2}{pi ^2+t^2}qquad (-frac pi 2 leq tleqfrac pi 2)$$ we get the cubic equation
      $$2 t^3-7 t^2+2 pi ^2 t-2 pi ^2=0$$ which has only one real root. Using the hyperbolic solution for one real root, the result is really ugly but it evaluates as $tapprox 1.428167$ that is to say $xapprox 0.714083$ while the "exact" solution is $xapprox 0.714836$.



      We can even do better building the $[2,2]$ Padé approximant around $x=frac pi 4$
      $$cos(2x)+4x-3=frac{(pi -3)+2 left(x-frac{pi }{4}right)+left(2-frac{2 pi }{3}right)
      left(x-frac{pi }{4}right)^2 } {1-frac{2}{3} left(x-frac{pi }{4}right)^2 }$$       Solving the quadratic
      $$x=frac pi 4+frac{6-sqrt{252-144 pi +24 pi ^2}}{4 pi -12}approx 0.714837$$



      To even avoid solving the quadratic equation, building instead the $[1,3]$ Padé approximant around $x=frac pi 4$ would lead to
      $$x=frac pi 4 -frac{(pi -3) left(15-6 pi +pi ^2right)}{4 left(12-6 pi +pi ^2right)}approx 0.714837$$






      share|cite|improve this answer































        0














        There is no solution in terms of elementary functions - you can solve it only by a numerical algorithm. There is Newton's method - the 'goto' method; but possibly there is an algorithm particular to this one that converges particularly fast. Numerical algorithms is a very highly developed area of mathematics, and for various problems numerical algorithms exist that converge truly astoundingly fast! The one I mentioned - Newtons - converges very fast at nearly every application of it ... and for the vast majority of problems you can do at least that well.






        share|cite|improve this answer





















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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5














          $$1le 1+sin^2(x)le 2implies frac 12le xle 1.$$
          the function
          $$f:xmapsto sin^2(x)+1-2x$$
          is continuous at $[frac 12,1],$



          $$f(frac 12)>0; ; f(1)<0,$$
          and



          $$f'(x)=sin(2x)-2<0.$$



          By IVT,
          there is a unique solution $alpha$ in $]frac 12,1[$.



          $$alpha=lim_{nto+infty}u_n$$



          with
          $$u_0=1$$
          and
          $$u_{n+1}=u_n-frac{f(u_n)}{f'(u_n)}$$



          This numerical method is known as Newton-Raphson.






          share|cite|improve this answer




























            5














            $$1le 1+sin^2(x)le 2implies frac 12le xle 1.$$
            the function
            $$f:xmapsto sin^2(x)+1-2x$$
            is continuous at $[frac 12,1],$



            $$f(frac 12)>0; ; f(1)<0,$$
            and



            $$f'(x)=sin(2x)-2<0.$$



            By IVT,
            there is a unique solution $alpha$ in $]frac 12,1[$.



            $$alpha=lim_{nto+infty}u_n$$



            with
            $$u_0=1$$
            and
            $$u_{n+1}=u_n-frac{f(u_n)}{f'(u_n)}$$



            This numerical method is known as Newton-Raphson.






            share|cite|improve this answer


























              5












              5








              5






              $$1le 1+sin^2(x)le 2implies frac 12le xle 1.$$
              the function
              $$f:xmapsto sin^2(x)+1-2x$$
              is continuous at $[frac 12,1],$



              $$f(frac 12)>0; ; f(1)<0,$$
              and



              $$f'(x)=sin(2x)-2<0.$$



              By IVT,
              there is a unique solution $alpha$ in $]frac 12,1[$.



              $$alpha=lim_{nto+infty}u_n$$



              with
              $$u_0=1$$
              and
              $$u_{n+1}=u_n-frac{f(u_n)}{f'(u_n)}$$



              This numerical method is known as Newton-Raphson.






              share|cite|improve this answer














              $$1le 1+sin^2(x)le 2implies frac 12le xle 1.$$
              the function
              $$f:xmapsto sin^2(x)+1-2x$$
              is continuous at $[frac 12,1],$



              $$f(frac 12)>0; ; f(1)<0,$$
              and



              $$f'(x)=sin(2x)-2<0.$$



              By IVT,
              there is a unique solution $alpha$ in $]frac 12,1[$.



              $$alpha=lim_{nto+infty}u_n$$



              with
              $$u_0=1$$
              and
              $$u_{n+1}=u_n-frac{f(u_n)}{f'(u_n)}$$



              This numerical method is known as Newton-Raphson.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 21 '18 at 17:19

























              answered Nov 21 '18 at 17:13









              hamam_Abdallah

              38k21634




              38k21634























                  2














                  Once it is proved that $f(x)=sin^2(x)-2x+1$ has a unique real zero in the interval $[0,pi/4]$, its numerical determination is simple since $f(x)$ is a positive and convex function on $(0,pi/4)$ (due to $f''(x)>0$), hence by applying Newton's method with starting point $x=0$ we get



                  $$ rho approx 0.714836$$
                  in just four steps.






                  share|cite|improve this answer


























                    2














                    Once it is proved that $f(x)=sin^2(x)-2x+1$ has a unique real zero in the interval $[0,pi/4]$, its numerical determination is simple since $f(x)$ is a positive and convex function on $(0,pi/4)$ (due to $f''(x)>0$), hence by applying Newton's method with starting point $x=0$ we get



                    $$ rho approx 0.714836$$
                    in just four steps.






                    share|cite|improve this answer
























                      2












                      2








                      2






                      Once it is proved that $f(x)=sin^2(x)-2x+1$ has a unique real zero in the interval $[0,pi/4]$, its numerical determination is simple since $f(x)$ is a positive and convex function on $(0,pi/4)$ (due to $f''(x)>0$), hence by applying Newton's method with starting point $x=0$ we get



                      $$ rho approx 0.714836$$
                      in just four steps.






                      share|cite|improve this answer












                      Once it is proved that $f(x)=sin^2(x)-2x+1$ has a unique real zero in the interval $[0,pi/4]$, its numerical determination is simple since $f(x)$ is a positive and convex function on $(0,pi/4)$ (due to $f''(x)>0$), hence by applying Newton's method with starting point $x=0$ we get



                      $$ rho approx 0.714836$$
                      in just four steps.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 21 '18 at 17:09









                      Jack D'Aurizio

                      287k33280658




                      287k33280658























                          1














                          Just for the fun of the approximation.



                          Using the double angle formula Rewrite the equation as $$cos(2x)+4x=3$$ Now, let $t=2x$ to make the equation
                          $$cos(t)+2t=3$$
                          Now, using the approximation
                          $$cos(t) simeqfrac{pi ^2-4t^2}{pi ^2+t^2}qquad (-frac pi 2 leq tleqfrac pi 2)$$ we get the cubic equation
                          $$2 t^3-7 t^2+2 pi ^2 t-2 pi ^2=0$$ which has only one real root. Using the hyperbolic solution for one real root, the result is really ugly but it evaluates as $tapprox 1.428167$ that is to say $xapprox 0.714083$ while the "exact" solution is $xapprox 0.714836$.



                          We can even do better building the $[2,2]$ Padé approximant around $x=frac pi 4$
                          $$cos(2x)+4x-3=frac{(pi -3)+2 left(x-frac{pi }{4}right)+left(2-frac{2 pi }{3}right)
                          left(x-frac{pi }{4}right)^2 } {1-frac{2}{3} left(x-frac{pi }{4}right)^2 }$$                           Solving the quadratic
                          $$x=frac pi 4+frac{6-sqrt{252-144 pi +24 pi ^2}}{4 pi -12}approx 0.714837$$



                          To even avoid solving the quadratic equation, building instead the $[1,3]$ Padé approximant around $x=frac pi 4$ would lead to
                          $$x=frac pi 4 -frac{(pi -3) left(15-6 pi +pi ^2right)}{4 left(12-6 pi +pi ^2right)}approx 0.714837$$






                          share|cite|improve this answer




























                            1














                            Just for the fun of the approximation.



                            Using the double angle formula Rewrite the equation as $$cos(2x)+4x=3$$ Now, let $t=2x$ to make the equation
                            $$cos(t)+2t=3$$
                            Now, using the approximation
                            $$cos(t) simeqfrac{pi ^2-4t^2}{pi ^2+t^2}qquad (-frac pi 2 leq tleqfrac pi 2)$$ we get the cubic equation
                            $$2 t^3-7 t^2+2 pi ^2 t-2 pi ^2=0$$ which has only one real root. Using the hyperbolic solution for one real root, the result is really ugly but it evaluates as $tapprox 1.428167$ that is to say $xapprox 0.714083$ while the "exact" solution is $xapprox 0.714836$.



                            We can even do better building the $[2,2]$ Padé approximant around $x=frac pi 4$
                            $$cos(2x)+4x-3=frac{(pi -3)+2 left(x-frac{pi }{4}right)+left(2-frac{2 pi }{3}right)
                            left(x-frac{pi }{4}right)^2 } {1-frac{2}{3} left(x-frac{pi }{4}right)^2 }$$                             Solving the quadratic
                            $$x=frac pi 4+frac{6-sqrt{252-144 pi +24 pi ^2}}{4 pi -12}approx 0.714837$$



                            To even avoid solving the quadratic equation, building instead the $[1,3]$ Padé approximant around $x=frac pi 4$ would lead to
                            $$x=frac pi 4 -frac{(pi -3) left(15-6 pi +pi ^2right)}{4 left(12-6 pi +pi ^2right)}approx 0.714837$$






                            share|cite|improve this answer


























                              1












                              1








                              1






                              Just for the fun of the approximation.



                              Using the double angle formula Rewrite the equation as $$cos(2x)+4x=3$$ Now, let $t=2x$ to make the equation
                              $$cos(t)+2t=3$$
                              Now, using the approximation
                              $$cos(t) simeqfrac{pi ^2-4t^2}{pi ^2+t^2}qquad (-frac pi 2 leq tleqfrac pi 2)$$ we get the cubic equation
                              $$2 t^3-7 t^2+2 pi ^2 t-2 pi ^2=0$$ which has only one real root. Using the hyperbolic solution for one real root, the result is really ugly but it evaluates as $tapprox 1.428167$ that is to say $xapprox 0.714083$ while the "exact" solution is $xapprox 0.714836$.



                              We can even do better building the $[2,2]$ Padé approximant around $x=frac pi 4$
                              $$cos(2x)+4x-3=frac{(pi -3)+2 left(x-frac{pi }{4}right)+left(2-frac{2 pi }{3}right)
                              left(x-frac{pi }{4}right)^2 } {1-frac{2}{3} left(x-frac{pi }{4}right)^2 }$$                               Solving the quadratic
                              $$x=frac pi 4+frac{6-sqrt{252-144 pi +24 pi ^2}}{4 pi -12}approx 0.714837$$



                              To even avoid solving the quadratic equation, building instead the $[1,3]$ Padé approximant around $x=frac pi 4$ would lead to
                              $$x=frac pi 4 -frac{(pi -3) left(15-6 pi +pi ^2right)}{4 left(12-6 pi +pi ^2right)}approx 0.714837$$






                              share|cite|improve this answer














                              Just for the fun of the approximation.



                              Using the double angle formula Rewrite the equation as $$cos(2x)+4x=3$$ Now, let $t=2x$ to make the equation
                              $$cos(t)+2t=3$$
                              Now, using the approximation
                              $$cos(t) simeqfrac{pi ^2-4t^2}{pi ^2+t^2}qquad (-frac pi 2 leq tleqfrac pi 2)$$ we get the cubic equation
                              $$2 t^3-7 t^2+2 pi ^2 t-2 pi ^2=0$$ which has only one real root. Using the hyperbolic solution for one real root, the result is really ugly but it evaluates as $tapprox 1.428167$ that is to say $xapprox 0.714083$ while the "exact" solution is $xapprox 0.714836$.



                              We can even do better building the $[2,2]$ Padé approximant around $x=frac pi 4$
                              $$cos(2x)+4x-3=frac{(pi -3)+2 left(x-frac{pi }{4}right)+left(2-frac{2 pi }{3}right)
                              left(x-frac{pi }{4}right)^2 } {1-frac{2}{3} left(x-frac{pi }{4}right)^2 }$$                               Solving the quadratic
                              $$x=frac pi 4+frac{6-sqrt{252-144 pi +24 pi ^2}}{4 pi -12}approx 0.714837$$



                              To even avoid solving the quadratic equation, building instead the $[1,3]$ Padé approximant around $x=frac pi 4$ would lead to
                              $$x=frac pi 4 -frac{(pi -3) left(15-6 pi +pi ^2right)}{4 left(12-6 pi +pi ^2right)}approx 0.714837$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Nov 23 '18 at 4:24

























                              answered Nov 22 '18 at 18:12









                              Claude Leibovici

                              119k1157132




                              119k1157132























                                  0














                                  There is no solution in terms of elementary functions - you can solve it only by a numerical algorithm. There is Newton's method - the 'goto' method; but possibly there is an algorithm particular to this one that converges particularly fast. Numerical algorithms is a very highly developed area of mathematics, and for various problems numerical algorithms exist that converge truly astoundingly fast! The one I mentioned - Newtons - converges very fast at nearly every application of it ... and for the vast majority of problems you can do at least that well.






                                  share|cite|improve this answer


























                                    0














                                    There is no solution in terms of elementary functions - you can solve it only by a numerical algorithm. There is Newton's method - the 'goto' method; but possibly there is an algorithm particular to this one that converges particularly fast. Numerical algorithms is a very highly developed area of mathematics, and for various problems numerical algorithms exist that converge truly astoundingly fast! The one I mentioned - Newtons - converges very fast at nearly every application of it ... and for the vast majority of problems you can do at least that well.






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      There is no solution in terms of elementary functions - you can solve it only by a numerical algorithm. There is Newton's method - the 'goto' method; but possibly there is an algorithm particular to this one that converges particularly fast. Numerical algorithms is a very highly developed area of mathematics, and for various problems numerical algorithms exist that converge truly astoundingly fast! The one I mentioned - Newtons - converges very fast at nearly every application of it ... and for the vast majority of problems you can do at least that well.






                                      share|cite|improve this answer












                                      There is no solution in terms of elementary functions - you can solve it only by a numerical algorithm. There is Newton's method - the 'goto' method; but possibly there is an algorithm particular to this one that converges particularly fast. Numerical algorithms is a very highly developed area of mathematics, and for various problems numerical algorithms exist that converge truly astoundingly fast! The one I mentioned - Newtons - converges very fast at nearly every application of it ... and for the vast majority of problems you can do at least that well.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 21 '18 at 17:24









                                      AmbretteOrrisey

                                      57410




                                      57410






























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