# Find the domain of f(x)=square root sin x + square root (16-x^2) .

Answers (1)

Solution:   We have ,   $f(x)=\sqrt{\sin x}+\sqrt{(16-x^{2})}$

For     $\sqrt{(16-x^{2})}\Rightarrow 16-x^{2}\geq 0$

$\Rightarrow$              $-4\leq x\leq 4$

For      $\sqrt{\sin x}\Rightarrow \sin x\geq 0\Rightarrow 2n\pi\leq x\leq (2n+1)\pi$

Thus , taking intersection with  $-4\leq x\leq 4$ we get $[-4,-\pi]\cup [0,\pi].$

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