8.227. $\sin t^{2}-\sin t=0$.

8.227. $\sin t^{2}-\sin t=0$.
## Solution.

Using the formula $\sin \alpha-\sin \beta=2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$, we get

$$
2 \sin \frac{t^{2}-t}{2} \cos \frac{t^{2}+t}{2}=0
$$

From this,

1) $\sin \frac{t^{2}-t}{2}=0$
2) $\cos \frac{t^{2}+t}{2}=0$

From the first equation, we have

$$
\frac{t^{2}-t}{2}=\pi k, k \in \mathbb{Z} \Leftrightarrow t^{2}-t-2 \pi k=0, k \in \mathbb{Z}
$$

Solving this quadratic equation, we get $t_{1,2}=\frac{1 \pm \sqrt{1+8 \pi k}}{2}$, where $1+8 \pi k \geq 0, k \geq-\frac{1}{8 \pi}, k=0,1,2, \ldots$.

From the second equation, we have

$$
\frac{t^{2}+t}{2}=\frac{\pi}{2}+\pi n, n \in \mathbb{Z}, \text{ or } t^{2}+t-\pi-2 \pi n=0, n \in \mathbb{Z}
$$

Solving this quadratic equation, we find $t_{3,4}=\frac{-1 \pm \sqrt{1+4 \pi(1+2 n)}}{2}$, where $1+4 \pi(1+2 n) \geq 0, n \geq-\frac{1}{8 \pi}-\frac{1}{2}, n=0,1,2, \ldots$.

Answer: $t_{1,2}=\frac{1 \pm \sqrt{1+8 \pi k}}{2} ; t_{3,4}=\frac{-1 \pm \sqrt{1+4 \pi(1+2 n)}}{2}$, where $k$ and $n=0,1,2, \ldots$.