The numbers of solutions of the pair of equations 2sin2⁡θ−cos⁡2θ=0, 2cos2⁡θ−3sin⁡θ=0 in the interval [0,2π] is

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zero

one

two

four

answer is C.

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Detailed Solution

2sin2⁡θ−cos⁡2θ=0⇒ 1−cos⁡2θ−cos⁡2θ=0⇒cos⁡2θ=1/2⇒ 2cos2⁡θ−1=1/2⇒2cos2⁡θ=3/2So that from 2cos2⁡θ−3sin⁡θ=0, we havesin⁡θ=1/2⇒θ=π/6,5π/6 as θ∈[0,2π].

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