Answer
$5\sqrt 2 \sin (2x+\dfrac{7\pi}{4})$
Work Step by Step
Here, we have $5 \sin 2x-5 \cos 2x=R \sin (2x-\phi)$
Since, the angle $\phi$ lies in the first quadrant so, $\phi=45^{\circ}$
Thus, $\cos \phi=\cos 45^{\circ}=\dfrac{\sqrt 2}{2}$; $\sin \phi=\sin 45^{\circ}=\dfrac{\sqrt 2}{2}$
Thus, $R^2(\cos^2 \phi+\sin^2 \phi)=(5)^2+(5)^2$
This gives: $R=5 \sqrt 2$
since, the trigonometric function has period of $2 \pi$, so we will have to add $2 \pi$
$5 \sin 2x-5 \cos 2x=R \sin (2x-\phi)$
Thus, we have $5 \sin 2x-5 \cos 2x=5 \sqrt 2\sin (2x-\dfrac{\pi}{4}+2\pi)=5\sqrt 2 \sin (2x+\dfrac{7\pi}{4})$
