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Given that $y=\sin(x)$ is an expicit function of the first-order differential equation $\frac{dy}{dx}=\sqrt{1-y^2}$. Find an interval I of definition, the solution interval.

So I got to the point that $\cos(x)={\cos(x)}$, what do I do next?

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  • $\begingroup$ For what values of $x$ is $\cos x = \sqrt{1 - \sin^2 x}$? This is not true for all real numbers $x$. $\endgroup$ – Simon S Jan 28 '15 at 23:59
  • $\begingroup$ Oh okay, so there would be no interval, right? $\endgroup$ – Julia Hall Jan 29 '15 at 0:00
  • $\begingroup$ No, think again. $\endgroup$ – Simon S Jan 29 '15 at 0:01
  • $\begingroup$ what are the constant solutions of this differential equation? $\endgroup$ – abel Jan 29 '15 at 0:04
  • $\begingroup$ @SimonS I really don't know. I just started this course. $\endgroup$ – Julia Hall Jan 29 '15 at 0:19
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$\sqrt{1-\sin ^2 x}\geq 0$. Hence $\cos x \geq 0$ which requires $0\leq x\leq \pm \pi/2$.

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  • $\begingroup$ How could ever $0\leq x \leq -\pi/2$? $\endgroup$ – Mårten W Aug 9 '16 at 0:20
  • $\begingroup$ That is a mistake. It is $0\leq |x| \leq \pi/2$. Thanks for pointing out. $\endgroup$ – Purushothaman Aug 10 '16 at 11:30

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