Q) Let S = {Let S = {θε [−2 π, 2π] : 2 cos2 θ + 3 sin θ = 0}.Then the sum of the elements of S is

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Let S={θ∈[-2π,2π]:2 〖cos〗^2⁡〖θ+3 sin⁡〖θ=0}〗 〗, then the sum of the elements of S is

Let S={0∈[-2π,2π]:2cos^2⁡〖θ+3 sin⁡〖θ=0}.〗 〗 Then the sum of the elements of S is:

Let S={θ∈[-2 π, 2 π]: 2 cos ^2θ+3 sinθ=0}. Then the sum of the elements of S is [JEE Main - 2019(...

let `theta in[-2pi,2pi]` and `2cos^(2)theta+3sintheta=0` then sum of all solution is

Let S={θ∈[-2π,2π] ∶2 cos^2⁡〖θ+3 sin⁡θ 〗=0}. Then the sum of the element of S is

The sum of all values of θ ∈[0,2π] satisfying 2(sin)^2 θ = cos2θ and 2(cos)^2 θ = 3sinθ is

Let S = {θ ∈ (0, 2π): 7 cos^2 θ - 3 sin^2 θ - 2 cos^2 2θ = 2}. Then, the sum of roots of all the equ

The sum of all values of θ ∈ [0, 2π] satisfying 2 sin^2θ = cos 2θ and 2 cos² θ = 3 sin sin θ is