For x∈(−π2,π2), find the solution set of the inequality |4sinx−1|<√5.
Given: sinπ10=√5+14
A
x∈(−3π10,3π10)
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B
x∈(−π10,π10)
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C
x∈(−π2,π10)
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D
x∈(−π10,3π10)
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Solution
The correct option is Dx∈(−π10,3π10) Given the inequality: |4sinx−1|<√5 ⇒−√5<4sinx−1<√5 ⇒−√5+14<sinx<√5+14 ⇒−sinπ10<sinx<cosπ5 ⇒sin(−π10)<sinx<sin(π2−π5) ⇒sin(−π10)<sinx<sin(3π10) ⇒x∈(−π10,3π10)