Question

Find x if sin inverse 6x + sin inverse 6 root 3 x equal to minus 5 by 2

Answer 1

Answer is: - 1/12

Step-by-step explanation :

$= > {sin}^{ - 1} 6x + {sin}^{ - 1} (6 \sqrt{3x} ) = \frac{ - \pi}{2} \\ = > { - sin }^{ - 1} 6x = \frac{\pi}{2} + {sin}^{ - 1}(6 \sqrt{3x} ) \\ = > taking \: sin \: on \: both \: sides \\ = > sin( \frac{\pi}{2} + {sin}^{ - 1}(6 \sqrt{3x} )) = sin({sin}^{ - 1}6x) \\ = > sin( \frac{\pi}{2} + {sin}^{ - 1}(6 \sqrt{3x} )) = - 6x \\ = > as \: sin( \frac{\pi}{2} + x) = cosx \\ = > so \: cos( {sin}^{ - 1} (6 \sqrt{3x}) = - 6x \\ = > as \: {sin}^{ - 1} a = {cos}^{ - 1} ( \sqrt{1 - {a}^{2} } ) \\ = > so \: cos( {cos}^{ - 1} \sqrt{1 - 108 {x}^{2} } ) = - 6x \\ = > \sqrt{1 - 108 {x}^{2} } = - 6x \\ = > squaring \: both \: side \\ = > 1 - 108 {x}^{2} = 36 {x}^{2} \\ = > 1 = 144 {x}^{2} \\ = > \frac{1}{144} = {x}^{2} \\ = > x = \frac{1}{12} and \: x = \frac{ - 1}{12} \\ = > but \: x = \frac{1}{12} will \: not \: satisfy \: the \: answer \\ = > therefore \: x = \frac{ - 1}{12} (ans)$

Hope you have got it