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Find the value of θ satisfying 11sin3θ-43cos2θ7-7-2=0

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a
2nπ±π6, n∈I
b
nπ+-1nπ6, n∈I
c
nπ+-1nπ3, n∈I
d
2nπ±π3, n∈I

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detailed solution

Correct option is B

We have 1     1    sin3θ-4   3    cos2θ7   -7       -2=0          ⇒ 0    1    sin3θ-7    3    cos2θ14  -7       -2 = 0                           ∵ C1→C1−C2         ⇒7 0    1    sin3θ-1    3    cos2θ 2   -7       -2 = 0                                                            taking 7 common from C1⇒70-1(2-2cos2θ)+sin3θ(7-6)=0                                                               expanding along R1⇒7[-2(1-cos2θ)+sin3θ]=0⇒-14+14cos2θ+7sin3θ=0⇒14cos2θ+7sin3θ=14⇒141-2sin2θ+73sinθ-4sin3θ=14⇒-28sin2θ+14+21sinθ-28sin3θ=14⇒-28sin2θ-28sin3θ+21sinθ=0⇒28sin3θ+28sin2θ-21sinθ=0⇒4sin3θ+4sin2θ-3sinθ=0⇒sinθ4sin2θ+4sinθ-3=0⇒  Either sinθ=0⇒θ=nπ or 4sin2θ+4sinθ-3=0⇒sinθ=-4±16+488=-4±648          =-4±88=48,-128⇒sinθ =12,−32⇒sinθ = 12 = sinπ6Thenθ = nπ+(-1)nπ6]


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