QJEEM19Ma0345: JEE Main 2019 Mathematics - Class 12 JEEM19Ma0345 Solution: If (x > ), then x is equal to :

Question

Verified Solution • 2025–26 Session moderate

If

{\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}

(x >

3 \over 4

), then x is equal to :

Accepted Answer

Given,

{\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}

\Rightarrow {\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) = {\pi \over 2} - {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right)

\Rightarrow \cos \left( {{{\cos }^{ - 1}}\left( {{2 \over {3x}}} \right)} \right) = \cos \left[ {{\pi \over 2} - {{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right]

\Rightarrow {2 \over {3x}} = \sin \left\{ {{{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right\}

\Rightarrow {2 \over {3x}} = \sin \left\{ {{{\sin }^{ - 1}}{{\sqrt {16{x^2} - 9} } \over {4x}}} \right\}

\Rightarrow {2 \over {3x}} = {{16{x^2} - 9} \over {4x}}

\Rightarrow 64 = 9\left( {16{x^2} - 9} \right)

\Rightarrow 16{x^2} - 9 = {{64} \over 9}

\Rightarrow 16{x^2} = {{64} \over 9} + 9

\Rightarrow 16{x^2} = - {{145} \over 9}

\Rightarrow x = \pm {{\sqrt {145} } \over {4 \times 3}}

\Rightarrow x = \pm {{\sqrt {145} } \over {12}}

as given that

x &gt; {3 \over 4}

\therefore

x

\ne

- {{\sqrt {145} } \over {12}}

\therefore

x

= {{\sqrt {145} } \over {12}}

Step-by-Step Verified Solution