X = c1 cos t + c2 sin t is a two-parameter family of solutions of the second-order DE x0 + x = 0. Find a solution of the second-order IVP consisting of this differential equation and the
given initial conditions.
1. x(pi/6)=1/2 , xprime(pi/6)=0

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)

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.