$x_1(t)=3cos(2t+pi/2)$ and $x_2(t)=4cos(2t)$ given. What is the result of $x(t)=x_1(t)+x_2(t)$?

When I do the calculations, I get the result as below.

$$x(t)=x_1(t)+x_2(t)=3cos(2t+pi/2)+x_2(t)=4cos(2t)$$

$$x(t)=3left(cos(2t)cos(pi/2)-sin(2t)sin(pi/2)right)+4cos(2t)$$

$$x(t)=-3sin(2t)+4cos(2t)$$

I wonder If I can simply the result in $x(t)=Acos(omega t + phi)$ form.