(a) 在 $0\leq x\leq 2\pi$ 上画出 $y=3\sin x+2$ 的图像草图。

(b) 分别确定下列方程在区间 $0\leq x\leq 2\pi$ 内的解的个数。

(i) $3\sin x+2=x$

(ii) $3\sin x+2=5-x$

(c) 解方程 $3\sin x+2=5\cos^2 x-1$，其中 $0\leq x\leq 2\pi$。

先画出正弦曲线的一整个周期。振幅是 $3$，中线是 $y=2$，因此最高点和最低点分别在 $y=5$ 和 $y=-1$。

The graph has period $2\pi$, amplitude $3$ and midline $y=2$. Mark

Join these points with a smooth sine curve, with horizontal tangents at the maximum and minimum.

在 (a) 的坐标系内画过原点的直线 $y=x$。方程的每个解对应直线与正弦曲线的一个交点。

The line $y=x$ crosses the sine curve once, between $\frac\pi2$ and $\pi$. Before $\frac\pi2$, the curve is at height at least $2$ while the line is below $2$. After $\pi$, the curve is at height at most $2$ while the line is above $2$. Hence there is one solution.

画出另一条直线 $y=5-x$。用 $(0,5)$ 和 $(5,0)$ 定位，再沿 $x$ 轴从左到右数交点。

On the same sketch, the line $y=5-x$ passes through $(0,5)$ and $(5,0)$. It crosses the curve three times, with one crossing in each of

Thus there are three solutions.

用 $\cos^2x=1-\sin^2x$ 把方程化为关于 $\sin x$ 的二次方程。正值对应第一、第二象限，$-1$ 对应一个最低点。

Replacing $\cos^2x$ by $1-\sin^2x$ gives

Thus $\sin x=\frac25$ or $\sin x=-1$. If $\alpha=\sin^{-1}(2/5)$, the first equation gives $x=\alpha$ or $\pi-\alpha$. The second gives $x=\frac{3\pi}2$. The solutions are