The set of all x in (−π2,π2) satisfying |4sinx−1|<√5 is given by
A
(−π10,3π10)
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B
(π10,−3π10)
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C
(π10,3π10)
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D
None of these
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Solution
The correct option is A(−π10,3π10) ∣4sinx−1∣<√5 ⇒−√5<4sinx−1<√5 ⇒−√5+14<sinx<√5+14 Hence solution in the given interval is, xϵ(−π10,3π10) Hence, option 'A' is correct.