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Question

For x(π2,π2), find the solution set of the inequality |4sinx1|<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 D x(π10,3π10)
Given the inequality: |4sinx1|<5
5<4sinx1<5
5+14<sinx<5+14
sinπ10<sinx<cosπ5
sin(π10)<sinx<sin(π2π5)
sin(π10)<sinx<sin(3π10)
x(π10,3π10)

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