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Inverse Trigonometric Functions question

2018 · Shift 1 · Q28
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  5. /2018 · Shift 1 · Q28

Inverse Trigonometric Functions question

2018 · Shift 1 · Q28

JEE AdvancedMathematicsInverse Trigonometric FunctionsNumerical+3 / −1
The number of real solutions of the equation sin⁡−1(∑i=1∞xi+1−x∑i=1∞(x2)i)=π2−cos⁡1(∑i=1∞(−x2)i−∑i=1∞(−x)i)\begin{aligned} & {\sin ^{ - 1}}\left( {\sum\limits_{i = 1}^\infty {} {x^{i + 1}} - x\sum\limits_{i = 1}^\infty {} {{\left( {{x \over 2}} \right)}^i}} \right) \\ & = {\pi \over 2} - {\cos ^1}\left( {\sum\limits_{i = 1}^\infty {} {{\left( {{{ - x} \over 2}} \right)}^i} - \sum\limits_{i = 1}^\infty {} {{\left( { - x} \right)}^i}} \right) \\\end{aligned}​sin−1(i=1∑∞​xi+1−xi=1∑∞​(2x​)i)=2π​−cos1(i=1∑∞​(2−x​)i−i=1∑∞​(−x)i)​ lying in the interval (−12,12)\left( { - {1 \over 2},{1 \over 2}} \right)(−21​,21​) is ........... . (Here, the inverse trigonometric functions sin −-− 1 x and cos −-− 1 x assume values in [−π2,π2]{\left[ { - {\pi \over 2},{\pi \over 2}} \right]}[−2π​,2π​] and [0,π]{\left[ {0,\pi } \right]}[0,π], respectively.)
Numerical answer
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