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

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)'

f'(x)=1-cosx

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:

f(0)=0-sin0=0-0=0.

**Because f(x) is an one-one function, x=0 is the only solution for the equation x-sinx=0.**

sinx = x

Solution

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 .

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.