Posted by **farouk** on Tuesday, November 15, 2011 at 3:53pm.

Show that the equation x+3sinx=2 has a root between 0.4 and 0.6.express the root to four significant figures

- calculus -
**MathMate**, Tuesday, November 15, 2011 at 4:43pm
First part:

let

f(x)=x+3sin(x)-2

f(0.4)=-0.432

f(0.6)=0.294

By the intermediate value theorem for a continuous function, there must be a value of x=x0 between 0.4 and 0.6 such that f(0.4)≤f(x0)≤f(0.6), specifically where f(x0)=0.0.

There are many ways to find the root of f(x)=0, for example, arrange the equation as:

x=arcsin((2-x0)/3)

and iterate starting with x0=0.5

then x=arcsin((2-0.5)/3)=0.5236

Let x0=0.5236 and continue until the value is stable.

Alternatively, use Newton's method for the iteration:

x1=x0-(x0+3sin(x0)-2)/(1+3cos(x0))

Starting from x0=0.5, you will get 9 digits of accuracy with 2 iterations.

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