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Integral value of x for which the equation 2{\cos ^{ - 1}}x = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) is satisfied is

Option: 1

-1


Option: 2

0


Option: 3

1


Option: 4

doesnt exist


Answers (1)

best_answer

Thus required equation became.

2{\cos ^{ - 1}}\left( {\cos y} \right) = {\sin ^{ - 1}}\sin 2{y} \to (1)

So  {\sin ^{ - 1}}(\sin 2y) = 2y        - \frac{\pi }{4} \le y \le \frac{\pi }{4}

And    2{\cos ^{ - 1}}(\cos y) = 2y             o \le y \le \pi

equation 1 holds only when 

y\epsilon [0,\frac{\pi}{4}]

x\epsilon [\frac{\sqrt2}{2},1]

Posted by

Divya Prakash Singh

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