how to solve this equation?

(1+cost)(tcost-sint)=2K(sint^2), where k is a constant

Answer 1

[Mohammad Abouali]

well a quick look at the equation, t=0 is always the solution regardless of the value of K.

However, if you want to solve these numerically you can choose an approach like below:

K=2;
fun=@(t) (1+cos(t)).*(t.*cos(t)-sin(t))-2*K*(sin(t).^2);
% or the following. Not sure which one you want.
% fun=@(t) (1+cos(t)).*(t.*cos(t)-sin(t))-2*K*(sin(t.^2));

Options=optimset(@fsolve);
Options.TolFun=1e-10;

[x,fval,exitflag,output] = fsolve(fun,1,Options)

x =

     1.4560e-04

fval =

    -8.4804e-08

exitflag =

       1

output = 

         iterations: 12
          funcCount: 26
          algorithm: 'trust-region-dogleg'
      firstorderopt: 9.8791e-11
            message: 'Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the selected valu...'

[Ankit agrawal]

many thanks for the solution ...can i plot this equation in a graph....in the domain(1, pi/2)

[Mohammad Abouali]

Sure you can.

K=2;
fun1=@(t) (1+cos(t)).*(t.*cos(t)-sin(t));
fun2=@(t) 2*K*(sin(t.^2));

ezplot(fun1,[1,pi])
hold on
h=ezplot(fun2,[1,pi]);
set(h,'Color','r')
legend('(1+cost)(tcost-sint)','2K(sint^2)')

where fun is your function defined in previous code.

BTW, if you are interested in the solution between the [1,pi] you can use fminbnd function as follow:

fminbnd(fun,1,pi)

ans =

      1.2915

fminbnd(fun,1.3,pi)

ans =

      2.7949