Differentiate the given solution:
[tex]x=C_1\cos(t)+C_2\sin(t) \implies x'=-C_1\sin(t)+C_2\cos(t)[/tex]
Now, given that x (π/4) = √2/2 … (I'm assuming there are symbols missing somewhere) … you have
[tex]\dfrac{\sqrt2}2=C_1\cos\left(\dfrac\pi4\right)+C_2\sin\left(\dfrac\pi4\right)[/tex]
[tex]\implies\dfrac1{\sqrt2} = \dfrac{C_1}{\sqrt2}+\dfrac{C_2}{\sqrt2}[/tex]
[tex]\implies C_1+C_2=1[/tex]
Similarly, given that x' (p/4) = 0, you have
[tex]0=-C_1\sin\left(\dfrac\pi4\right)+C_2\cos\left(\dfrac\pi4\right)[/tex]
[tex]\implies 0=-\dfrac{C_1}{\sqrt2}+\dfrac{C_2}{\sqrt2}[/tex]
[tex]\implies C_1=C_2[/tex]
From this result, it follows that
[tex]C_1+C_2=2C_1=1 \implies C_1=C_2=\dfrac12[/tex]
So the particular solution to the DE that satisfies the given conditions is
[tex]\boxed{x=\dfrac12\cos(t)+\dfrac12\sin(t)}[/tex]
