solution:
we have the solution as
x(t)= c1*cos(t) + c2*sin(t)
=>x'(t) = -c1*sin(t) +c2*cos(t)
given x(π/3)=sqrt(3)/2 and x'(π/3)=0
=>sqrt(3)/2= c1*(1/2) + c2*(sqrt(3)/2) ------------------------> (1)
and 0= -c1*(sqrt(3)/2) + c2*(1/2) --------------------------->(2)
=>c2=c1*sqrt(3)
using this in (1) we get
sqrt(3)/2= c1*(1/2) + c1*(3/2)
=>c1= sqrt(3)/4
and c2=3/4
x(t) = (sqrt(3)/4)*cos(t) +(3/4)*sin(t)
