Answer
$x =\frac{ 5\sqrt{13}}{2} sin ( 2πt + tan^{−1}( \frac{1}{2\sqrt{3}}))$
Refer to the graph below that confirms that the two equations of $x$ are the same
Work Step by Step
Let $ω = 2π$. Then $A sin(ωt + θ) = A(cos θ sin 2πt + sin θ cos 2πt) = (A cos θ) sin 2πt + (A sin θ) cos 2πt$
so for the two equations for $x$ to be equivalent, we need $A cos θ = 5\sqrt{3}$ and $A sin θ = \frac{5}{2}$.
These imply that $A^2 = (A cos θ)^2 + (A sin θ)^2 = \frac{325}{4}$ and $tan θ = \frac{A sin θ}{ A cos θ} = \frac{1} {2√3}$.
So let $A = \sqrt{\frac{325}{4}} = \frac{5\sqrt{13}}{2}$ and $θ = tan^{−1} (\frac{1}{2√3})$.
Then $cos θ = \frac{2\sqrt{3}} {\sqrt{13}}$ and $sin θ = \frac{1}{\sqrt{13}}$, so $A cos θ = 5\sqrt{3}$ and $A sin θ = \frac{5}{2}$, as required.
Hence $x =\frac{ 5\sqrt{13}}{2} sin ( 2πt + tan^{−1}( \frac{1}{2\sqrt{3}}))$
