[tex]a\sin(b(x-h))=a\sin(bx)\cos(bh)-a\cos(bx)\sin(bh)[/tex]
For this to be equivalent to [tex]5\cos x+2\sin x[/tex], you require [tex]b=1[/tex] and
[tex]\begin{cases}a\cos h=2\\-a\sin h=5\end{cases}[/tex]
Dividing the second equation by the first gives
[tex]\dfrac{-a\sin h}{a\cos h}=-\tan h=\dfrac52\implies h=\arctan\left(-\dfrac52\right)=-\arctan\dfrac52[/tex]
Meanwhile, you also get
[tex]a\cos\left(-\arctan\dfrac52\right)=2\implies a\cos\left(\arctan\dfrac52\right)=\dfrac{2a}{\sqrt{29}}=2\implies a=\sqrt{29}[/tex]
So,
[tex]f(x)=5\cos x+2\sin x=\sqrt{29}\sin\left(x+\arctan\dfrac52\right)[/tex]
