I was working on a problem booklet and came across the following equation. $$\sqrt2\sin(2x)-\cos(2x)=\sqrt3\sin(2x-a)$$
$a \in \mathbb{R}$ is a specific value that I'm supposed to find, but I don't see how to make the first part look like the second part in the first place, so I can't even get there
The standard way to do this is to write $$ \sqrt2\sin(2x)-\cos(2x)=A\sin2x\cos a-A\cos2x\sin a $$ with $A>0$, because the addition formula says the expression is $$ A\sin2x\cos a-A\cos2x\sin a=A\sin(2x-a). $$ We can choose $$ A\cos a=\sqrt{2},\quad A\sin a=1 $$ that gives $$ A^2=A^2\cos^2a+A^2\sin^2a=2+1=3. $$ Therefore $A=\sqrt{3}$ and so $$ \cos a=\frac{\sqrt{2}}{\sqrt{3}},\quad \sin a=\frac{1}{\sqrt{3}}. $$ Since both sine and cosine are positive, you know that you can take $0<a<\pi/2$, thus $$ a=\arcsin\frac{1}{\sqrt{3}}. $$