<s>[INST] 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 $A,B,C$.
 [/INST] Given a non obtuse triangle $ABC$, 
find the corresponding angles $\alpha,\beta,\gamma$
such that
\begin{align} 
\tfrac12\cos2\alpha+\sqrt2\cos\beta+\sqrt2\cos\gamma
&=\tfrac32
\tag{1}\label{1}
\\
\text{or }\quad
\cos^2\alpha+\sqrt2\cos\beta+\sqrt2\cos\gamma
-2&=0
.
\tag{1a}\label{1a}
\end{align}  
Using a known identity for triangles
\begin{align} 
\cos\alpha+\cos\beta+\cos\gamma
&=1+\frac{r}R
\tag{2}\label{2}
,
\end{align}
where $r$ and $R$ are the radii 
of inscribed and circumscribed circles, respectively,
we have 
\begin{align} 
\sqrt2\cos\alpha+\sqrt2\cos\beta+\sqrt2\cos\gamma
&=\sqrt2\,\left(1+\frac{r}R \right)
\tag{3}\label{3}
.
\end{align}
Let's introduce a parameter
$v=\frac{r}R$. It is also known that 
for valid triangles $0<v\le \tfrac12$.
Equations \eqref{1a},\eqref{2} provide
a quadratic equation in $\cos\alpha$
in terms of $v$:
\begin{align}
\cos^2\alpha
-\sqrt2\cos\alpha+\sqrt2(1+v)-2
&=0
\tag{4}\label{4}
.
\end{align}
The two solutions of \eqref{4} are
\begin{align}
s_+&=\tfrac{\sqrt2}2
+\tfrac12\sqrt{10-4\sqrt2\,(v+1)}
,\\
s_-&=\tfrac{\sqrt2}2
-\tfrac12\sqrt{10-4\sqrt2\,(v+1)}
.
\end{align}
Since $s_+>1$ $\forall v\in[0,\tfrac12]$,
the only valid solution is
\begin{align}
\cos\alpha&=\tfrac{\sqrt2}2
-\tfrac12\sqrt{10-4\sqrt2\,(v+1)}
\tag{5}\label{5}
.
\end{align}
The equation \eqref{5} shortens the valid range of $v$
to $[\sqrt2-1.\tfrac12]$,
since corresponding $\cos\alpha<0$ 
$\forall v\in(0,\sqrt2-1)$.
With the help of
\begin{align}
\cos\gamma&=
\cos(\pi-\alpha-\beta)
=-\cos(\alpha+\beta)
=-\cos\alpha\cos\beta+\sin\alpha\sin\beta
,
\end{align}
\eqref{2} can be transformed to
\begin{align}
\sin\alpha\sin\beta
&=1+v-\cos\alpha-\cos\beta
+\cos\alpha\cos\beta
,\\
\sin^2\alpha\sin^2\beta
&=(1+v-\cos\alpha-\cos\beta+\cos\alpha\cos\beta)^2
,\\
(1-\cos^2\alpha)(1-\cos^2\beta)
&=
(\cos\alpha-1)^2\cos^2\beta
\\
&+2(\cos\alpha-1)(1+v-\cos\alpha)\cos\beta
\\
&+(1+v-\cos\alpha)^2
,\\
\end{align} 
and we have a quadratic equation for $\cos\beta$
\begin{align}
\cos^2\beta
-(1+v-\cos\alpha)\cos\beta
-\cos\alpha
+v-\frac{v^2}{2(\cos\alpha-1)}
\tag{6}\label{6}
,
\end{align} 
The discriminant of \eqref{6} is
\begin{align}
\Delta&=\tfrac14\,\frac{(\cos\alpha+1)(\cos\alpha^2-2\cos\alpha \,v-1+2v+v^2)}{(\cos\alpha-1)^3}
,
\end{align}
condition $\Delta\ge0$
can be simplified to
\begin{align}
\frac{(\cos\alpha^2-2\cos\alpha \,v-1+2v+v^2)}{(\cos\alpha-1)^3}
&\ge0
,\\
(\cos\alpha^2-2\cos\alpha \,v-1+2v+v^2)
&\le0
,
\end{align}
which we can simplify by substitution 
$\cos^2\alpha=2+\sqrt2\cos\alpha-\sqrt2(1+v)$ as
\begin{align}
(\sqrt2-2v)\cos\alpha+1+v^2+(2-\sqrt2)v-\sqrt2
&\le0
,\\
(\sqrt2-2v)
\tfrac{\sqrt2}2
-\tfrac12\sqrt{10-4\sqrt2\,(v+1)}
+1+v^2+(2-\sqrt2)v-\sqrt2
&\le0
\tag{7}\label{7}
.
\end{align}
Considering $v\in[0,\tfrac12]$,
it can be shown that \eqref{7} holds for $v\in[0,\sqrt2-1]$
(rigorous proof is left to anyone interested),
that is,
\begin{align}
D(v)&=0,\quad \text{for }\quad v=\sqrt2-1
,\\
D(v)&<0,\quad \text{for }\quad v>\sqrt2-1
.
\end{align}

Thus, the only valid solution of \eqref{1}
is restricted to $v=\sqrt2-1$,
\begin{align}
\alpha&=\tfrac\pi2,\quad
\beta=\gamma=\tfrac\pi4
.
\end{align}
A 3D illustration - the surfaces
given in implicit form
\begin{align} 
\tfrac12\cos2x+\sqrt2\cos y+\sqrt2\cos z
-\tfrac32&=0
,\\
x+y+z-\pi&=0
,\\
x,y,z&\in[0,\tfrac\pi2]
\end{align}
shows only one intersection point, shown in yellow:

</s>