Question #3ded9

1 Answer
Wataru
Apr 1, 2017

Let #f(x)=x-cos x#.

We know that #f# is continuous on #[0,pi/2]# since it is the difference of two continuous functions #x# and #cos x#, and

#f(0)=-1<0< pi/2=f(pi/2)#.

By Intermediate Value Theorem, there exists #c in (0,pi/2)# s.t.
#f(c)=c-cos(c)=0#, which means that #c=cos(c)#.

Hence, #x=cos x# has a solution #c in (0,pi/2)#.

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