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Show that the equation x=a sinx+ b where 0 0, has at least one positive root.

```
6 years ago

```							Consider the case when a & b are strictly positive.Let f(x)=x-a.sinx-b . f(0)=0-a.0-b=-b<0.
and f(2.pi.([a]+1) + b) = 2.pi.([a]+1) +b - a.sinb - b = 2.pi.([a]+1) - a.sinb = x (say), where [a] denotes the greatest positive integer less than or equal to a.
now, as a>0 and sinb<= 1{<= means less than or equal to &  >= means greater than or equal to}, we have x=> 2.pi.([a]+1) -a >0 as 2.pi>4.
so as the values of f at 0 and at another positive number have different signs so it must have a positive root.
This technique is an application of the Intermediate Value Theorem (look up google if you want) for continuous functions.
```
6 years ago
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