The interval for which {sin ^{ - 1}}sqrt x + | vedclass

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(C) We know that the identity ${\sin ^{ - 1}} 	heta + {\cos ^{ - 1}} 	heta = \frac{\pi }{2}$ is valid for all $	heta \in [-1, 1]$.In this problem,$

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$[0, 1]$

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(C) We know that the identity ${\sin ^{ - 1}} 	heta + {\cos ^{ - 1}} 	heta = \frac{\pi }{2}$ is valid for all $	heta \in [-1, 1]$.In this problem,$	heta = \sqrt x$.For $\sqrt x$ to be defined,we must have $x \ge 0$.For the inverse trigonometric functions to be defined,we must have $-1 \le \sqrt x \le 1$.Since $\sqrt x$ is always non-negative,the condition $-1 \le \sqrt x \le 1$ simplifies to $0 \le \sqrt x \le 1$.Squaring the inequality,we get $0^2 \le x \le 1^2$,which results in $0 \le x \le 1$.Therefore,the interval for which the equation holds is $x \in [0, 1]$.

The interval for which ${\sin ^{ - 1}}\sqrt x + {\cos ^{ - 1}}\sqrt x = \frac{\pi }{2}$ holds is:

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