The domain of definition of the function $$fx=$$$$sin$$−$$1$$⁡$$(2x)+$$π$$6for$$ $$real-valued$$ x is

Question

The domain of definition of the function $$fx=$$$$sin$$−$$1$$⁡$$(2x)+$$π$$6for$$ $$real-valued$$ x is

Infinity Learn · Accepted Answer

For $$f(x)=$$$$sin$$−$$1$$⁡(2x)+$$π$$6$$ to be defined and real, $$sin$$−$$1$$⁡$$2x+$$π$$/6$$≥$$0$$ or  $$sin$$−$$1$$⁡$$2x$$≥−π$$6-----(1) But we know that −π$$/2$$≤$$sin$$−$$1$$⁡$$2x$$≤π$$/2$$ …. (2)Combining$$ $$(l) and (2),  −π$$/6$$≤$$sin$$−$$1$$⁡$$2x$$≤π$$/2$$ or  $$sin$$⁡$$($$−π$$/6)$$≤$$2x$$≤$$sin$$⁡$$($$π$$/2)$$ or  −$$1/2$$≤$$2x$$≤$$1$$ or  −$$1/4$$≤x≤$$1/2$$∴Domain of the function $$=$$−$$14$$,$$12$$

$T h e d o m a i n o f d e f i n i t i o n o f t h e f u n c t i o n f (x) = \sqrt{\sin^{- 1} (2 x) + \frac{π}{6}}$ $f o r r e a l - v a l u e d x i s$

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The domain of definition of the function fx=sin−1⁡(2x)+π6for real-valued x is

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$T h e d o m a i n o f d e f i n i t i o n o f t h e f u n c t i o n f (x) = \sqrt{\sin^{- 1} (2 x) + \frac{π}{6}}$ $f o r r e a l - v a l u e d x i s$