Question

class 12 maths.. Inverse trigonometrypls send this​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

Solve for x :-

$\rm :\longmapsto\: {sin}^{ - 1}x + {sin}^{ - 1}2x = \dfrac{\pi}{3}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: {sin}^{ - 1}x + {sin}^{ - 1}2x = \dfrac{\pi}{3}$

$\rm :\longmapsto\: {sin}^{ - 1}2x = \dfrac{\pi}{3} - {sin}^{ - 1}x$

$\rm :\longmapsto\: 2x = sin \bigg( \dfrac{\pi}{3} - {sin}^{ - 1}x \bigg)$

$\rm :\longmapsto\:2x = sin\dfrac{\pi}{3}cos( {sin}^{ - 1}x) - sin( {sin}^{ - 1}x) \: cos\dfrac{\pi}{3}$

$\green{\boxed{ \bf \because \: sin(x + y) = sinx \: cosy \: + \: cosx \: siny}}$

$\rm :\longmapsto\:2x = \dfrac{ \sqrt{3} }{2} \: cos( {cos}^{ - 1} \sqrt{1 - {x}^{2} } ) - x\: \times \dfrac{1}{2}$

$\green{\boxed{ \bf \because \: {sin}^{ - }x ={cos}^{-1} \sqrt{1-{x}^{2}}}}$

$\rm :\longmapsto\:2x + \dfrac{x}{2} = \dfrac{ \sqrt{3} }{2} \:\sqrt{1 - {x}^{2} } )$

$\rm :\longmapsto\:\dfrac{4x + x}{2} = \dfrac{ \sqrt{3} }{2} \:\sqrt{1 - {x}^{2} } )$

$\rm :\longmapsto\:\dfrac{5x}{2} = \dfrac{ \sqrt{3} }{2} \:\sqrt{1 - {x}^{2} } )$

$\rm :\longmapsto\:5x = \sqrt{3} \times \sqrt{1 - {x}^{2} }$

On squaring both sides, we get

$\rm :\longmapsto\: {25x}^{2} = 3(1 - {x}^{2})$

$\rm :\longmapsto\: {25x}^{2} = 3 - 3{x}^{2}$

$\rm :\longmapsto\: {25x}^{2} + 3{x}^{2} = 3$

$\rm :\longmapsto\: {28x}^{2} = 3$

$\rm :\longmapsto\: {x}^{2} = \dfrac{3}{28}$

$\bf\implies \:x = \: \pm \: \dfrac{ \sqrt{3} }{2 \sqrt{7} }$

Verification :-

Case :- 1

$\rm :\longmapsto\:When \: x = \dfrac{ \sqrt{3} }{2 \sqrt{7} }$

Consider, LHS

$\rm :\longmapsto\: {sin}^{ - 1}x + {sin}^{ - 1}2x$

$\rm \: = \: \: {sin}^{ - 1} \dfrac{ \sqrt{3} }{2 \sqrt{7} } + {sin}^{ - 1}\dfrac{ \sqrt{3} }{\sqrt{7} }$

$\rm \: = \: \: {tan}^{ - 1} \dfrac{ \sqrt{3} }{5} + {tan}^{ - 1}\dfrac{ \sqrt{3} }{2 }$

$\rm \: = \: \: {tan}^{ - 1} \bigg(\dfrac{\dfrac{ \sqrt{3} }{5} + \dfrac{ \sqrt{3} }{2} }{1 - \dfrac{ \sqrt{3} }{5} \times \dfrac{ \sqrt{3} }{2} } \bigg)$

$\rm \: = \: \: {tan}^{ - 1} \bigg(\dfrac{7 \sqrt{3} }{7 } \bigg)$

$\rm \: = \: \: {tan}^{ - 1}( \sqrt{3)}$

$\rm \: = \: \: \dfrac{\pi}{3}$

$\bf\implies \: LHS = RHS$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \green{\boxed{ \quad\bf \: x = \dfrac{ \sqrt{3} }{2 \sqrt{7} } \quad \: }}$

Case :- 2

$\rm :\longmapsto\:When \: x = - \: \dfrac{ \sqrt{3} }{2 \sqrt{7} }$

Consider LHS

$\rm :\longmapsto\: {sin}^{ - 1}x + {sin}^{ - 1}2x$

$\rm \: = \: \: {sin}^{ - 1} \dfrac{ - \sqrt{3} }{2 \sqrt{7} } + {sin}^{ - 1}\dfrac{ - \sqrt{3} }{\sqrt{7} }$

$\rm \: = \: \: - {sin}^{ - 1} \dfrac{ \sqrt{3} }{2 \sqrt{7} } - {sin}^{ - 1}\dfrac{ \sqrt{3} }{\sqrt{7} }$

$\rm \: = \: \: - {tan}^{ - 1} \dfrac{ \sqrt{3} }{5} - {tan}^{ - 1}\dfrac{ \sqrt{3} }{2 }$

$\rm \: = \: \: - {tan}^{ - 1} \bigg(\dfrac{\dfrac{ \sqrt{3} }{5} + \dfrac{ \sqrt{3} }{2} }{1 - \dfrac{ \sqrt{3} }{5} \times \dfrac{ \sqrt{3} }{2} } \bigg)$

$\rm \: = \: \: - {tan}^{ - 1} \bigg(\dfrac{7 \sqrt{3} }{7 } \bigg)$

$\rm \: = \: \: - {tan}^{ - 1}( \sqrt{3)}$

$\rm \: = \: \: - \: \dfrac{\pi}{3}$

$\bf\implies \: LHS \: \ne \: RHS$

So Solution of

$\bf :\longmapsto\: {sin}^{ - 1}x + {sin}^{ - 1}2x = \dfrac{\pi}{3} \: \: is \: \: x \: = \:\dfrac{ \sqrt{3} }{2 \sqrt{7} }$