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Question

# If α is a root of equation (2sinx−cosx)(1+cosx)=sin2x, β is a root of equation 3cos2x−10cosx+3=0 and γ is a root of equation 1−sin2x=cosx−sinx; 0≤α,β,γ≤π2, then sin(α−β) is equal to

A
166
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B
1236
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C
1266
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D
3226
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Solution

## The correct option is C 1−2√66(2sinx−cosx)(1+cosx)=sin2x⇒(2sinx−cosx)(1+cosx)=1−cos2x⇒(1+cosx)(2sinx−cosx−1+cosx)=0⇒(1+cosx)(2sinx−1)=0 ⇒cosx=−1 or sinx=12 Since, 0≤α≤π2 ⇒sinα=12, cosα=√32 ...(1) 3cos2x−10cosx+3=0⇒(3cosx−1)(cosx−3)=0⇒cosx=13 (∵cosx∈[−1,1])∴cosβ=13, sinβ=2√23 ...(2) Now, sin(α−β)=sinαcosβ−cosαsinβ=12×13−√32×2√23=1−2√66

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