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)#.

Related questions
See all questions in Intemediate Value Theorem
Impact of this question
2757 views around the world
You can reuse this answer
Creative Commons License