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Example

Consider the equation:

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

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:

\begin{displaymath}
\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
\end{displaymath}

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 $\pi/4=0.7853981635$.



Jean-Pierre Merlet 2012-12-20