Consider the cubic equation x3−(1+cosθ+sinθ)x2+(cosθsinθ+cosθ+sinθ)x−sinθcosθ=0 whose roots are x1,x2,x3 The number of values of θ in [0,2π] for which at least two roots are equal
A
3
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B
4
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C
5
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D
6
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Solution
The correct option is C5 Roots are sinθ,cosθ and 1 When sinθ=1, 2 equal roots are sinθ and 1 When cosθ=1. 2 equal roots are cosθ and 1 When sinθ=cosθ, 2 equal roots are sinθ and cosθ Hence θ=0,π2,2π,π4,5π4