A $0 \leq t \leq \pi$ real parameter for which values does the equation $\sin (x+t)=1-\sin x$ have no solution?
Transform the equation as follows:

$$
\sin \left[\left(x+\frac{t}{2}\right)+\frac{t}{2}\right]+\sin \left[\left(x+\frac{t}{2}\right)-\frac{t}{2}\right]=1
$$

According to the known identity for the sine of the sum and difference of angles, the equation can be further transformed as:

$$
\sin \left(x+\frac{t}{2}\right) \cos \frac{t}{2}+\cos \left(x+\frac{t}{2}\right) \sin \frac{t}{2}+\sin \left(x+\frac{t}{2}\right) \cos \frac{t}{2}-\cos \left(x+\frac{t}{2}\right) \sin \frac{t}{2}=1
$$

From this,

$$
2 \sin \left(x+\frac{t}{2}\right) \cos \frac{t}{2}=1
$$

The equation has no solution if $\cos \frac{t}{2}=0$, that is, if $t=\pi$.

If $t \neq \pi$, rearrange the equation:

$$
\sin \left(x+\frac{t}{2}\right)=\frac{1}{2 \cdot \cos \frac{t}{2}}
$$

According to the condition, now $0 \leq \frac{t}{2}<\frac{\pi}{2}$, so the right side is positive and $\sin \left(x+\frac{t}{2}\right) \leq 1$; therefore, there is a solution exactly when

$$
2 \cos \frac{t}{2} \geq 1
$$

From this, $\cos \frac{t}{2} \geq \frac{1}{2}$, which holds on the given interval if $\frac{t}{2} \leq \frac{\pi}{3}$. In this case, $\frac{1}{2} \leq \frac{1}{2 \cos \frac{t}{2}} \leq 1$ and for such values, there is an $x$ such that $\sin \left(x+\frac{t}{2}\right)=\frac{1}{2 \cos \frac{t}{2}}$.

Therefore, the equation has no solution exactly when $\frac{2 \pi}{3}<t \leq \pi$.