Riemann surface
matlab

Example
Suppose we wish to solve the equation $\cos x=-0.5$ and we look for all solutions lying in the interval $0 \leq x \leq 360^{\circ}$.

As before we start by looking at the graph of $\cos x$. This is shown in Figure 2. We have drawn a dotted horizontal line where $\cos x=-0.5$. The solutions of the equation correspond to the points where this line intersects the curve. One fact we do know from the Table on page 2 is that $\cos 60^{\circ}=+0.5$. This is indicated on the graph. We can then make use of the symmetry to deduce that the first angle with a cosine equal to -0.5 is $120^{\circ}$. This is because the angle must be the same distance to the right of $90^{\circ}$ that $60^{\circ}$ is to the left. From the graph we see, from consideration of the symmetry, that the remaining solution we seek is $240^{\circ}$. Thus
$$x=120^{\circ}, 240^{\circ}$$

Example
Suppose we wish to solve $\sin 2 x=\frac{\sqrt{3}}{2}$ for $0 \leq x \leq 360^{\circ}$.
Note that in this case we have the sine of a multiple angle, $2 x$.
To enable us to cope with the multiple angle we shall consider a new variable $u$ where $u=2 x$, so the problem becomes that of solving
$$\sin u=\frac{\sqrt{3}}{2} \quad \text { for } 0 \leq u \leq 720^{\circ}$$
We draw a graph of $\sin u$ over this interval as shown in Figure 3 .

E-mail: help-assignment@gmail.com  微信:shuxuejun

help-assignment™是一个服务全球中国留学生的专业代写公司