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If sin^2 θ – 2sinθ – 1 = 0 is to be satisfied for exactly 4 distinct values of θ ∈ [0, nπ], ∀ n∈N ; then the least value of n is

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If sin2θ – 2sinθ – 1 = 0 is to be satisfied for exactly 4 distinct values of θ ∈ [0, nπ], ∀ n∈N ; then the least value of n is

(a) 2

(b) 6

(c) 4

(d) 8

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Correct option is (c) 4

clearly four solutions lie in [0,4π] 

\(\therefore\) least value of n is 4

Also 5th solution lies in [0,5π] 

\(\therefore\) greatest value can be 5.

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