NOTE:  THIS VERSION HAS CORRECTIONS MADE ON 8 SEPTEMBER; NOTE MY EARLIER
POSTED f(t) WAS OF THE OPPOSITE SIGN, AND I ZOOMED-IN ON THE WRONG INTERVAL

GOAL: minimize 
  d(t) = (cos(t) + 5) - (sin(2.1 t) + 2)    on  0 <= t <= 2 pi
NOTE:
  d'(t) = f(t) = - sin(t) - 2.1 cos(2.1 t)
SO WE SOLVE
  f(t) = 0

BELOW:  FIRST DO SOME GRAPHING.
THEN WE APPROXIMATE f(t) BY LINE l(t) WHICH GOES THROUGH
  (3.5,f(3.5)),  (3.8,f(3.8))
AND WE WRITE THE LINE IN FORM
  l(t) = A + B (t - 3.5)
SO THE APPROXIMATE SOLUTION IS
  t = 3.5 - A / B


bueler@bueler-pogo:~$ octave

octave:1> t=0:.001:2*pi;
octave:2> f=-sin(t)-2.1*cos(2.1*t);
octave:3> plot(t,f)
octave:4> grid on
octave:5> t=3.5:.0001:3.8;
octave:6> f=-sin(t)-2.1*cos(2.1*t);
octave:7> plot(t,f), grid on
octave:8> t1=3.5;  A=-sin(t1)-2.1*cos(2.1*t1)
A = -0.66334
octave:9> t2=3.8;  B=-sin(t2)-2.1*cos(2.1*t2);  B = (B - A) / 0.3
B =  5.1305
octave:10> 3.5 - A / B
ans =  3.6293
octave:11> format long
octave:12> 3.5 - A / B
ans =  3.62929459640979