Question #3ded9

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Question #3ded9

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Answer 1 · 2017-04-01T19:10:14

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

We know that $f$ $f$ is continuous on $[0, \frac{π}{2}]$ $[0,pi/2]$ since it is the difference of two continuous functions $x$ $x$ and $cos x$ $cos x$ , and

$f (0) = - 1 < 0 < \frac{π}{2} = f (\frac{π}{2})$ $f(0)=-1<0< pi/2=f(pi/2)$ .

By Intermediate Value Theorem, there exists $c \in (0, \frac{π}{2})$ $c in (0,pi/2)$ s.t.
$f (c) = c - cos (c) = 0$ $f(c)=c-cos(c)=0$ , which means that $c = cos (c)$ $c=cos(c)$ .

Hence, $x = cos x$ $x=cos x$ has a solution $c \in (0, \frac{π}{2})$ $c in (0,pi/2)$ .

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Question #3ded9

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