Let S = θ ∈ [-2π, 2π]: 2cos2θ + 3sinθ = 0. Then the sum of the elements of S is

Question

Let S = θ ∈ [-2π, 2π]: 2cos2θ + 3sinθ = 0. Then the sum of the elements of S is (A) (13 π/6) (B) π (C) 2 π (D) (5 π/3)

Tardigrade Admin · Accepted Answer

2 (1- sin2 θ) + 3sin θ = 0 ⇒ 2 sin2 θ - 3 sin θ - 2 = 0 ⇒ (2 sin θ + 1 ) (sin θ - 2) = 0 ⇒ sin θ = - (1/2) ; sin θ = 2 (reject) roots: π + (π/6), 2π - (π/6), - (π/6), - π + (π/6) ⇒ sum of values = 2π

Q. Let $S$ = { $θ \in [- 2 π, 2 π] : 2 co s^{2} θ + 3 s in θ$ = 0}. Then the sum of the elements of $S$ is

$\frac{13 π}{6}$

$π$

$2 π$

$\frac{5 π}{3}$

Q. Let S = {θ∈[−2π,2π]:2cos2θ+3sinθ = 0}. Then the sum of the elements of S is

613π​

π

2π

35π​

2(1−sin2θ)+3sinθ=0 ⇒2sin2θ−3sinθ−2=0 ⇒(2sinθ+1)(sinθ−2)=0 ⇒sinθ=−21​;sinθ=2(reject) roots : π+6π​,2π−6π​,−6π​,−π+6π​ ⇒ sum of values = 2π

Q. Let $S$ = { $θ \in [- 2 π, 2 π] : 2 co s^{2} θ + 3 s in θ$ = 0}. Then the sum of the elements of $S$ is

$\frac{13 π}{6}$

$π$

$2 π$

$\frac{5 π}{3}$