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Question

If 1+sinx+sin2x+=4+23,0<x<π and xπ2, then x is equal to

A
π3,5π6
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B
2π2,π6
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C
π3,2π3
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D
π6,π3
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Solution

The correct option is C π3,2π3
Given, 1+sinx+sin2x+....=4+23
11sinx=4+23
1sinx=14+23=4234
sinx=32
x=π3,2π3[x(0,π)]

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