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Question

If ∣ ∣sin3θ11cos2θ43277∣ ∣=0, then find the number of values of θ in [0,2π].

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Solution

∣ ∣sin3θ11cos2θ43277∣ ∣C2C2+C3

∣ ∣sin3θ01cos2θ732147∣ ∣C2C2/7

7∣ ∣sin3θ01cos2θ13227∣ ∣
7(sin3θ(76))+1(2cos2θ2)
7(sin3θ(1))+2(cos2θ1)2(2(cos2θ1))
7(3sinθ7sin3θ)+4(sin2θ)
28sin3θ4sin2θ+21sinθ
=sinθ[28sin2θ+4sinθx]=0
sinθ=0
or 28sin2θ+4sinθ21=0
sinθ=237±114
or θ=0
and θ=sin1(237±114)

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