I ma supposed to find the domain of $$f:y=\sqrt{\sin \sqrt{x}} $$ I know that the first step is that: $$\sin \sqrt{x}\geq 0$$ and $$x\geq 0$$ When I plot the graph I know that $$\sin \sqrt{x}$$ is $$\geq 0$$ for $$\left [ 0,\infty \right ]$$, so does for x. Is it the domain then $$\left [ 0,\infty \right ]$$? Thanks
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- $\begingroup$ How did you determine $\sin\sqrt x\geq0$ for all $x\geq0$? For example, what if $x=(3\pi/2)^2$? $\endgroup$Clayton– Clayton2019-09-29 19:31:56 +00:00Commented Sep 29, 2019 at 19:31
- $\begingroup$ @Clayton I thought is it is from definition of te square root $\endgroup$Christine K.– Christine K.2019-09-29 19:35:36 +00:00Commented Sep 29, 2019 at 19:35
- $\begingroup$ did you not use the definition of $\sin(t)$? When is $\sin(t)\ge0$. Yes if $0\le t\le\pi$ but not if $\pi< t<2\pi$. Could you extend this, for a complete solution? $\endgroup$Mirko– Mirko2019-09-29 20:48:56 +00:00Commented Sep 29, 2019 at 20:48
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1 $$sin a\geq 0 \to 0+2k\pi\leq a\leq \pi+2k\pi$$so you put $\sqrt x $ in it to find domain. more than $x\geq 0 $ $$\\ sin \sqrt x \geq 0 \to 0+2k\pi\leq \sqrt x\leq \pi+2k\pi$$at the end note that $k$ must be in $\{0,1,2,3,...\}$
can you take over ?
- $\begingroup$ We need also $k>0$. $\endgroup$user– user2019-09-29 19:56:38 +00:00Commented Sep 29, 2019 at 19:56