Question

Find the domains of definition of the following functions.
$y=\sqrt{\sin x}+\sqrt{16-x^2}.$

Answer 1

Domain of $y=\sqrt{\sin x}+\sqrt{16-x^2}$

As $\sin x$ and $16-x^2$ is under root it should be greater than zero.

Thus $\sin x>0$ and $16-x^2>0$

$\sin x$ is greater than $0$ when $xϵ[0,\pi ]$ and $x^2<16$

thus $xϵ[0,\pi ]$ and $x<4$

Thus the domain of $y$ is $xϵ[0,1]$