How to use phasor algebra to solve for capacitive reactance?
This is how I solved, taking cos as reference to phasor equations:
$$V(t) = 230 \sin(\omega t+\pi/4)$$
$$I(t)=10\sin(\omega t - \pi/6)V$$
let  \$(\omega t+\pi/4) = \phi\$

we know:
$$sin\phi = -cos(\pi/2+\phi)$$
$$\implies V(t) = -230cos(\omega t+3\pi/4)$$
$$ I(t) = -10cos(\omega t+\pi/3)$$
Therefore in phasor form, it V and I can be represented as:
$$V = -230 e^{j3\pi/4}$$ 
$$I = -10e^{j\pi/3}$$
As everything is in phasor form, now we can apply ohm's law directly:

$$230e^{j3\pi/4} = (5 + 8j + X_{C})(10e^{j\pi/3})$$
$$\implies 23 e^{j5\pi/12} =  (5 + 8j + X_{C}) $$
$$\implies X_c = 23 cos(5\pi/12) + j 23 sin(5\pi/12) -5 - 8j$$
$$= 5.95 + 22.21 j - 5 -8j$$
$$= 0.95 + 14.21j$$
This will satisfy the given conditions in the question. As we can see the obtained expression for \$X_c\$ has a real part or resistive part and a positive imaginary part or inductive reactance part. Pure capacitance will only have negative imaginary part, i.e, with phasor -90 degrees.