Find the $3$ angles of triangle $ABC$ We have a non obtuse triangle $ABC$. 
With $$\bf\dfrac{1}{2}\cos(2A)+\sqrt{2}\cos(B)+\sqrt{2}\cos(C)=\dfrac{3}{2}$$
Find the $3$ angles $A,B,C$.

I would say this question doesn't really want you to go into any complicated algebra, etc. but rather just wants you to think. The left hand side involves factors of $\sqrt{2}$ but the right hand side is rational. It makes sense that these factors of $\sqrt{2}$ must disappear somehow. One possibility is that they "cancel out" entirely (i.e. $\cos(B)=-\cos(C)$ and $\cos(2A)=3$) but we see that this isn't possible. The other likely possibility is that "cancel out" through multiplication or division (i.e. $\cos(B)$ and $\cos(C)$ involve some factor of $\sqrt{2}$. This is probably a good time to say that I am writing very loosely in terms of mathematics.
It stands therefore that we should have $\cos(B)=\cos(C)=\frac{1}{\sqrt{2}}$ and hence $B=C=\frac{\pi}{4}$. Hence we would have $A=\frac{\pi}{2}$ for the sum of the internal angles of the triangle to be $\pi$ and we see that indeed, this is a solution.
