phasors and differential equation circuit analysis?
Thevnin voltage gives the open circuit voltage across two terminals (capacitor removed in this case).  And\$V_{th} = \cos(4t)\$ in here. 



When a load (capacitor in this case) is connected across these terminals, the voltage across the terminals A-B will change and you can use voltage division rule to obtain the output voltage. $$V_{AB} = V_{th}\times\dfrac{X_C}{R_{th}+X_C}$$ \$X_C\$-capacitive reactance.

The phasor method used you have used actually does that. So the answer is \$0.5\cos(4t-\pi/3)\$.