# Solve the equation x - sinx = 0.

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This is a transcedental equation and for solving it we have to differentiate the function.

f'(x)=1-cosx

The derivative is a monotone increasing function

( -1<cosx<1), so the difference 1 -cos x>0 =>f(x)>0, so f(x) is an injection.

We'll calculate f(0):

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

Because f(x) is an injection, that means that f(x) is increasing, so, x=0 is the only solution for x-sinx=0.

To solve x-sinx = 0.

Solution:

We know that sinx < x always for all x

So sinx /x < 1 always.

But Lt x--> 0 sinx/x = 1 is a well known limit.

Therefore sinx/x = 1 has no other solution than x=0.

x-sinx =0

==> sinx = x

If you assume that there are two functions:

y=sinx and y=x

The graph will show that both functions only intersect at one point (0,0)

Then the only solution is x=0

To check

sin 0 = 0

The only value for the equality is 0

sin 0 = 0

OR:

x-sinx > 0 for x>0

x-sinx <0 for x<0

x-sinx=0 for x=0