Here is a solution by the method of characteristics. So, we assume that the equation is
$$
u_t+u_x=\cos^2 u,\quad u(x,0)=u_0(x).
$$
Consider the characteristics defined by
$$
\frac{dx}{ds}=1,\quad x(0)=\tau,\\
\frac{dt}{ds}=1,\quad t(0)=0,\\
\frac{du}{ds}=\cos^2 u,\quad u(0)=u_0(\tau).
$$
Obviously, from first two equations
$$
x=s+\tau,\quad t=s\implies \tau=x-t.
$$
From the third equation
$$
\tan u=s+\tan u_0(\tau),
$$
or, finally,
$$
u=\tan^{-1}(\tan u_0(x-t)+t)
$$
as required.
Modifying the problem. Rather than consider the PDE $u_{xt}+u\ u_{xx}+\frac{1}{2}u_{x}^2=0$ with initial condition $u(x,0)=u_0(x)$ as asked above, I will consider the following variant.
$$
\text{Solve }u_{xy}+u\ u_{xx} + u_x^2=0\text{ subject to }u(x,0)=f(x).\qquad(\star)
$$
There are three differences between this question and that which was asked originally.
- The coefficient of $u_x^2$ has changed from $\frac{1}{2}$ to $1$.
- The variable $t$ has been renamed to $y$.
- The initial function $u_0(x)$ has been renamed to $f(x)$.
Only (1) represents a significant modification of the problem. It makes the solution more tractable and enables it to be found using an elementary application of the method of characteristics. For these reasons, it is conceivable that this was the intended question.
Note: I will not delve into regularity of the solutions in this answer.
Reduction to a first order quasilinear PDE. Write the equation as
$$
\frac{\partial}{\partial x}\left(u_y+u\ u_x\right)=0.
$$
Thus $(\star)$ is equivalent to
$$
u_y+u\ u_x=g(y),\qquad u(x,0)=f(x),\qquad (\star\star)
$$
where $g(y)$ is an arbitrary function of $y$ (with sufficient regularity).
Method of characteristics.
Perhaps the simplest formulation of the method of characteristics is for quasilinear first order PDEs. These are PDEs of the form
$$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u).$$
To solve this equation, one regards the solution as a surface $z=u(x,y)$ in $xyz$-space. Let $s$ parametrize the initial curve $\bigl(s,0,f(s)\bigr)$ and let $t$ be a second parameter, which can be thought of as the distance flowed along a characteristic curve emanating from $\bigl(s,0,f(s)\bigr)$.
The characteristic equations are then
$$
\frac{dx}{dt}=a(x,y,z),\quad \frac{dy}{dt}=b(x,y,z),\quad \frac{dz}{dt}=c(x,y,z).
$$
Returning to our equation $(\star\star)$, this reduces to $a(x,y,u)=u$ and $b(x,y,u)=1$ and $c(x,y,u)=g(y)$.
Thus
$$
\frac{dx}{dt}=z,\quad \frac{dy}{dt}=1,\quad \frac{dz}{dt}=g(y)
$$
with initial conditions $x(0)=s$ and $y(0)=0$ and $z(0)=f(s)$.
The solution to this system is
$$
x=s+zt,\quad y=t,\quad z=f(s)+h(t),
$$
where $h(t)$ is the antiderivative of $g(t)$ satisfying $h(0)=0$. Since $g$ was arbitrary, so is $h$ given $h(0)=0$.
The solution. Now we eliminate all occurrences of $t$ by replacing them with $y$, then eliminate $s$ by writing $s=x-zy$. Finally, replace $z$ with $u$ to obtain the implicit equation
$$
\boxed{u=f(x-uy)+h(y)},
$$
where $h(y)$ is any sufficiently regular function satisfying $h(0)=0$. This is an implicit equation for the general solution of $(\star)$.
TL;DR. Change the $\frac{1}{2}$ in the original question to $1$ to obtain a PDE solvable by the method of characteristics.
Best Answer
The point is, your function is a constant in respect to $s$, but not in respect to $r$, set $s=t+x, r=t-x$, if you do this you will get again $\frac{du}{ds}=0$, but integrating this you will get $u(s,r)=c(r)\implies u(x,t)=c(t-x)$ for some arbitrary function $c$.
Now we look on the condition: $\forall x,t\quad x^2+t^2=1\implies u(x,t)=c(t-x)=x$, we want to show that no such function exists, to do so we just need to find $x_0,t_0$ and $x_1,t_1$ such that $x_1\ne x_0$ and $t_0-x_0=t_1-x_1$.
Take $x_0=y_0=\frac{1}{\sqrt2}$, $x_1=y_1=-\frac{1}{\sqrt2}$, from the first tuple we get $c(0)=\frac{1}{\sqrt2}$ and from the second we get $c(0)=-\frac{1}{\sqrt2}$, thus $c$ is ill defined