Let S=θ∈[−2π,2π]:2cos2⁡θ+3sin⁡θ=0Then the sum of the elements of S is :

A
$2 π$
B
$π$
C
$5 π / 3$
D
$13 π / 6$

Solution:

Given, $2 \cos^{2} θ + 3 \sin θ = 0$

$\begin{array}{l} \Rightarrow 2 (1 - \sin^{2} θ) + 3 \sin θ = 0 \\ \Rightarrow 2 - 2 \sin^{2} θ + 3 \sin θ = 0 \\ \Rightarrow 2 \sin^{2} θ - 4 \sin θ + \sin θ - 2 = 0 \\ \Rightarrow 2 \sin θ (\sin θ - 2) + 1 (\sin θ - 2) = 0 \\ \Rightarrow (2 \sin θ + 1) (\sin θ - 2) = 0 \end{array}$

$\Rightarrow \sin θ = - \frac{1}{2}$ $[∵ \sin θ - 2 \neq 0]$

$\Rightarrow θ = \frac{- π}{6}, \frac{- 5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}$

Required sum $= [\frac{- π - 5 π + 7 π + 11 π}{6}] = \frac{12 π}{6} = 2 π$

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

Solution:

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