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**Hint:**To solve this question, we need to use the angle sum property of a triangle according to which the sum of all the angles of a triangle is $\pi $. A trigonometric equation typically has several solutions or an infinite number of solutions because all trigonometric ratios are periodic. Then, using trigonometric tan techniques, we will simplify the given equation to obtain the final equation.

**Formula Used**: The trigonometric formulas are:

$\sin (-\theta )=-\sin \theta $

**Complete step-by-step solution:**A trigonometric equation will also have a generic solution, which is stated in a generalized form in terms of "n" and contains all the values that would fulfill the given equation. As per the $\sin \theta $ formula, the sine of an angle $\theta $, in a right-angled triangle is equal to the ratio of the opposite side and hypotenuse.

The mostly applied value of $\theta $ satisfying both the equations

$\sin \theta =\dfrac{1}{2}$

We will take the angle sum property of the triangle and derive the equation of the angle. Hence, the most general value of θ satisfying the equations $\sin \theta \text{=}\sin \alpha $ and $\text{cos}\theta \text{=cos}\alpha $ is $\theta =n\pi +\alpha $.

$\begin{align}

& 2{{\sin }^{2}}\theta -3\sin \theta -2=0 \\

& \Rightarrow (2\sin \theta +1)(\sin \theta -2)=0 \\

& \Rightarrow \left| \sin \theta =\sin \left( \dfrac{-\pi }{6} \right) \right| \\

& \Rightarrow \theta =n\pi +{{(-1)}^{n}}\left( \dfrac{-\pi }{6} \right) \\

\end{align}$

$\begin{align}

& \Rightarrow \theta =n\pi +{{(-1)}^{n+1}}\left( \dfrac{\pi }{6} \right) \\

& \Rightarrow \theta =n\pi +{{(-1)}^{n}}\left( \dfrac{7\pi }{6} \right) \\

\end{align}$

**Option ‘D’ is correct**

**Note:**The sine of the angle divided by the cosine of that angle is known as the trigonometric ratio. It can be characterized as the proportion of the perpendicular side to the neighboring side of a right-angled triangle's sides. A solution to an ‘n-order’ ordinary differential equation that uses precisely ‘n’ necessary arbitrary constants.

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