Example

Consider the equation:

$$\left (\sin(t)-1/2\right )\left (\sin(t)-1/2\,\sqrt {3}\right )\left ( \cos(t)-1/2\,\sqrt {2}\right )=0$$

which has as roots: $\pi/6 (0.5235)$, $\pi/4 (0.7853)$, $\pi/3 (1.047)$, $2\pi/3 (2.094)$, $5\pi/6 (2.617)$, $7\pi/4 (5.497)$. This equation is equivalent to:

$$\sin^2(t)\cos(t)- 0.707106781\,\sin^2(t)- 1.366025404\,\sin(... ...s(t)+ 0.9659258263\,\sin(t)+ 0.4330127\,\cos(t)- 0.3061862179$$

The A, SSin, CCos vectors have the following values:

A	SSin	CCos
1	2	1
-0.7071067810	2	0
-1.366025404	1	1
0.9659258263	1	0
0.4330127020	0	1
-0.3061862179	0	0

If Inf=0 and Sup=0.8 the procedure indicates that there are two roots corresponding to $\pi/6=0.5235987758$ and $\pi/4=0.7853981635$.