Where does the equation in the blue circle come from?$]=>D_3=\frac{\sqrt[\phantom{2}]{2}{A}_1}{3}3cos{}_1-\frac{2A_2}{3}3cos{}_2=03=\sqrt[\phantom{2}]{2}{A}_{1}cos{}_1-2A_{2}cos{}_2=0=(3)$ new $[0$

$\{\begin{array}{c}\text{:}=>x_1(t)=\frac{A_1}{3}sin(\sqrt[\phantom{2}]{2}t+{}_1)-\frac{A_2}{3}sin(2t+{}_2)\end{array}$