Solving trigonometric equations numerically?

[Philipp Gressly Freimann]

Hello

I want to solve the following equation numerically between -PI and PI:

sin(x) = 0.5x + 0.2

[which is similar to sin(x) - 0.5x - 0.2 = 0]

The graph shows me three solutions. Is there a possibility to solve this equation numerically using sympy?

Thanks in advance

φ

[Oscar Benjamin]

You can use nsolve to find numerical solutions:
```
In [10]: nsolve(Eq(sin(x), 0.5*x + 0.2), x, 0)
Out[10]: 0.425436108484597
```
This will find one root at a time starting from an initial guess (I've
used zero).

Initial guesses -1 and +1 give two other roots.
```
In [11]: nsolve(Eq(sin(x), 0.5*x + 0.2), x, -1)
Out[11]: -2.11307244875263

In [12]: nsolve(Eq(sin(x), 0.5*x + 0.2), x, +1)
Out[12]: 1.59919364642736
```

You can get more precise solutions using prec:
```
In [15]: nsolve(Eq(sin(x), 0.5*x + 0.2), x, 0, prec=50)
Out[15]: 0.42543610848459725447179186114511470949330179080539
```

--
Oscar

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[Philipp Gressly Freimann]

Hello

Well, thanks a lot. Works great. I did not know the "nsolve" command.

If I am right, there is no command to find all three solutions?

φ

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[Oscar Benjamin]

There is no function to find all 3 solutions. It would be good to have
one. In general it can be hard even to know how many solutions there
are from a pure numerical algorithm but I think that something based
on interval-Newton would be useful:

https://en.wikipedia.org/wiki/Newton%27s_method#Interval_Newton's_method
https://en.wikipedia.org/wiki/Interval_arithmetic#Interval_Newton_method

On Sat, 21 Dec 2019 at 08:17, Philipp Gressly Freimann

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[Chris Smith]

Another approach for these problems is to use, as a starting point, solutions to a simpler problem which become initial guess to the more difficult problem. e.g. solving sin(x)=0.2 will give you two solutions and as many others as you want by adding or subtraction 2pi. Then these approximate solutions can be used as initial guesses for nsolve as you change the problem to `sin(x)-(a*x+0.2)` with `a` increasing as quickly as possible from 0 to 0.5 (in your case). This is the "continuation" method.

/c

[Philipp Gressly Freimann]

Thanks.