The general value of θ satisfying the equation 2sin⁡2θ−3sin⁡θ−2=0, is

A
$n π + (- 1)^{n} \frac{π}{6}$
B
$n π + (- 1)^{n} \frac{π}{2}$
C
$n π + (- 1)^{n} \frac{5 π}{6}$
D
$n π + (- 1)^{n} \frac{7 π}{6}$

Solution:

We have,

$\begin{array}{l} 2 \sin^{2} θ - 3 \sin θ - 2 = 0 \\ \Rightarrow (2 \sin θ + 1) (\sin θ - 2) = 0 \\ \Rightarrow \sin θ = - \frac{1}{2} \\ \Rightarrow \sin θ = \sin (- \frac{π}{6}) \\ \Rightarrow θ = n π + (- 1)^{n} (- \frac{π}{6}), n \in Z \\ \Rightarrow θ = n π + (- 1)^{n + 1} \frac{π}{6}, n \in Z \end{array}$

The general value of $θ$ satisfying the equation $2 \sin 2 θ - 3 \sin θ - 2 = 0$ , is

Solution:

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