Differentiating the general solution
x(t) = C₁ cos(t) + C₂ sin(t)
gives
x'(t) = -C₁ sin(t) + C₂ cos(t)
Given the initial conditions x(π/6) = 1/2 and x'(π/6) = 0, we have
1/2 = C₁ cos(π/6) + C₂ sin(π/6) = √3/2 C₁ + 1/2 C₂
===> 1 = √3 C₁ + C₂
and
0 = -C₁ sin(π/6) + C₂ cos(π/6) = -1/2 C₁ + √3/2 C₂
===> 0 = C₁ - √3 C₂
Solve for C₁ and C₂. The second equation says C₁ = √3 C₂, and substituting this into the first equation gives
1 = √3 (√3 C₂) + C₂
1 = 3 C₂ + C₂
1 = 4 C₂
C₂ = 1/4
and it follows that
C₁ = √3/4
So, the particular solution to the DE is
x(t) = 1/4 cos(t) + √3/4 sin(t)