cos^2x-sin^2=cos2x，

So M=[0, 1]

x- 1/i=(ix- 1)/i

Turn fractions into common denominators

=(-x-i)/(- 1)

Multiply up and down by I, I 2 =-1.

=x+i

Therefore, the modulus of (x- 1/i is the radical number x 2+1

(Solution of module: module of a+bi = radical A 2+B 2)

Square of both sides x 2+1< 2

Answer:-1

So choose C.

Method 2: special value method

Take endpoint 0, 1.

Obviously 0, 1 belongs to m.

Substitute 0, 1 into x-1/i.

Find the modulus and compare it with the root number 2.