Question

Sin-1(2-3x square) find the domain

Answer 1

Answer:

[-1 , 1]

Step-by-step explanation:

Range of Sin is from -1 to 1

Range of Sin inverse is - pi/2 to pi/2

$- \frac{ \pi}{2} < { \sin}^{ - 1} (2 - 3 {x}^{2} ) < \frac{ \pi}{2} \\ \\ apply \: sin \\ \\ \sin( - \frac{ \pi}{2} ) < \sin( { \sin}^{ - 1}(2 - 3 {x}^{2} )) < \sin(\frac{ \pi}{2}) \\ \\ - 1 < 2 - 3 {x}^{2} < 1 \\ \\ multiply \: \: - 1 \\ \\ - 1 < 3 {x}^{2} - 2 < 1 \\ \\ 1 < 3 {x}^{2} < 3 \\ \\ \frac{1}{3} < {x}^{2} < 1 \\ \\ since \: domain \: is \: in \: numerator \\ \\ < = \leq \\ \\ {x}^{2} - \frac{1}{3} \geq0 \\ \\ this \: does \: not \: satisfy \: being \: domain \\ \\ therefore \\ \\ domain \: is \: {x}^{2} - 1 \leq0$

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