It is possible to solve the equation x = sin x?

4 Answers

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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


Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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 .








thewriter's profile pic

thewriter | College Teacher | (Level 1) Valedictorian

Posted on

sin x can be expressed as a power series which is

sinx= x-(x^3/3!)+(x^5/5!)-(x^7/7!)+(x^9/9!)...

If x=six x,

then x-(x^3/3!)+(x^5/5!)-(x^7/7!)+(x^9/9!)...=x

==> -(x^3/3!)+(x^5/5!)-(x^7/7!)+(x^9/9!)...=0

==> x^5/5!+x^9/9!...=x^3/3!+x^7/7!...

This is possible only if x=0, as then the terms on both the sides are equal to 0.