Answer
$\sqrt {29} \sin (3t-68^{\circ} )$
Work Step by Step
The given equation has form of: $A\sin (Bt+\phi)$ where, $A=\sqrt {a_1^2+a_2^2}$ We are given that $-2 \sin (3t )+ 5\cos(3 t)$ with $a_1=-2$ and $a_2=5$ and $B=3$ So, we have: $A=\sqrt {(-2)^2+(5)^2}=\sqrt {29}$
Now, $\cos \phi =\dfrac{a_1}{A}=-\dfrac{2}{\sqrt {29}}; \sin \phi =\dfrac{a_2}{A}=\dfrac{5}{\sqrt {29}}$ and $\tan \phi =\dfrac{a_2}{a_1}=\dfrac{-5}{2}$
Now, $\phi =\tan^{-1}(\dfrac{-5}{2})= -68^{\circ}$ and
Thus, $\phi =-68^{\circ}$
Therefore, we have: $-2 \sin (3t )+ 5\cos(3 t)=A\sin (Bt+\phi)=\sqrt {29} \sin (3t-68^{\circ} )$
