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Q.

One of the general solutions of 3cosθ3sinθ=4sin2θcos3θ is 

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a

nπ+π/18,nZ

b

nπ/2+π/6,nZ

c

nπ/3+π/18,nZ

d

nπ3+π9,nZ

answer is C.

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Detailed Solution

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 We have 3cosθ3sinθ=2(sin5θsinθ)

orcos(θ+π/6)=sin5θ=cos(π/25θ)

θ=(nπ/3)+(π/18)

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One of the general solutions of 3cos⁡θ−3sin⁡θ=4sin⁡2θcos⁡3θ is