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 sinα+sinβ+sinγ can be equal to
A
14−3√26√2
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B
56
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C
3+4√26
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D
1+√22
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Solution
The correct option is C3+4√26 (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)