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It is possible to solve the equation x = sin x?
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Yes, it is possible to solve the equation, using derivative of the function associated to the expression:
f(x) = x - sinx
The function is a continuous function, formed by elementary functions as the linear one,x , and the trigonometrical one, sin x, so we can calculate it's derivative.
f'(x) = (x - sinx)'
We notice that the first derivative is an increasing function.
( -1<cosx<1), so the difference 1 -cos x>0 =>f(x)>0, so f(x) is one-one function.
We can also do a very simple calculus:
Because f(x) is an one-one function, x=0 is the only solution for the equation x-sinx=0.
Posted by giorgiana1976 on August 18, 2010 at 12:45 AM (Answer #1)
High School Teacher
You can solve it by graphing both functions.
Let y = x
and g = sinx
We need to determine in which both functions intersect .
let us determin y values;
x= -1 ==> y= -1
x=0 ==> y= 0
x = 1==> y= 1
Now for g:
We know that sinx values are between -1, and 1
Then the only intersecting point is when x = 0
==> sinx = x
==> sin0 =0
==< 0 = 0
Posted by hala718 on August 18, 2010 at 12:50 AM (Answer #2)
sin x can be expressed as a power series which is
If x=six x,
This is possible only if x=0, as then the terms on both the sides are equal to 0.
Posted by thewriter on August 18, 2010 at 12:56 AM (Answer #3)
High School Teacher
sinx = x
Let P be a point on a circle with centre making an angle x with the initial OX line and radius r.. Le OX cut the circle at P'. Let PA be perpendicular to OX to meet OX at A.
Then, the area of OAP < Area OPP' < area OPX, where XP is the tangent at P. Therefore,
rsinx < rx < rtanx. Divide by rsinx.
1 < x/sinx < 1/cosx
cosx < sinx/x < 1.
So sinx < x always. But in limit cosx < sinx/x as x--> 0 becomes 1 < sinx/x.
So sinx/x = 1 only inlimit when x--> 0 when sinx = x.
So x = 0 .
Posted by neela on August 18, 2010 at 1:34 AM (Answer #4)
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