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Question

# Complete the set of values of x in (0,π) satisfying the inequation 1+log2sinx+log2sin3x≥0 is

A
[0,π6]
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B
[π2,3π4]
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C
[π6,π4][3π4,5π6]
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D
[5π6,π]
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Solution

## The correct option is C [π6,π4]∪[3π4,5π6] x∈(0,π)⇒3x∈(0,3π) For the inequality to be defined, sinx>0 and sin3x>0 ⇒x∈(0,π3)∪(2π3,π) ⋯(1) 1+log2sinx+log2sin3x≥0 ⇒log2(2sinxsin3x)≥0 ⇒2sinxsin3x≥1 ⇒2sinx(3sinx−4sin3x)≥1 ⇒8sin4x−6sin2x+1≤0 Let sin2x=t Then 8t2−6t+1≤0 ⇒(t−12)(t−14)≤0 ⇒14≤t≤12 ⇒14≤sin2x≤12 ⇒12≤sinx≤1√2 ⇒x∈[π6,π4]∪[3π4,5π6] ⋯(2) From (1) and (2), x∈[π6,π4]∪[3π4,5π6]

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