当 $0<k<1$ 时，水平线 $y=k$ 在 $0<x\leqslant\pi$ 内会与 $y=\sin x$ 相交两次：一次在第一象限，一次在第二象限。

For $0<k<1$, the equation

has one solution in $0<x<\dfrac{\pi}{2}$ and one solution in $\dfrac{\pi}{2}<x<\pi$.

Therefore the number of solutions is

$\tan x$ 的周期是 $\pi$，在 $0<x<\pi$ 内只有第一象限那一次取正值 $k$。在第二象限，$\tan x<0$。

Since $0<k<1$, $\tan x=k$ has one solution in the first quadrant.

On $\dfrac{\pi}{2}<x<\pi$, $\tan x<0$, so there are no further solutions.

Therefore the number of solutions is

令 $u=2x$，则区间变成 $0<u\leqslant200\pi$。$\cos u=k$ 在每个完整的 $2\pi$ 周期内有两个解，而 $0$ 到 $200\pi$ 一共有 $100$ 个周期。

Let

As $0<x\leqslant100\pi$,

For $0<k<1$, the equation

has two solutions in each interval of length $2\pi$.

There are $100$ complete periods in $0<u\leqslant200\pi$, so the number of solutions is

因为 $0<k<1$，所以 $1<k+1<2$。但 $\sin x$ 的最大值是 $1$，不可能等于一个大于 $1$ 的数。

Since

we have

But

Therefore $\sin x=k+1$ has no solutions.

The number of solutions is