Animated Solution for Mathematics - Trigonometry: The general values of $\theta$ satisfying the equation $2{\sin }^2\theta -\sin \theta -2=0$ is

The given equation is $2{\sin }^2\theta -3\sin \theta -2=0$. To solve this, we treat it as a quadratic equation by substituting $x=\sin \theta$.

This transformation yields the standard quadratic form:

To solve $2x^2-3x-2=0$, we look for two numbers that multiply to $2\times (-2)=-4$ and add to $-3$. These numbers are $-4$ and $1$.

Splitting the middle term, we rewrite the equation as:

Factoring by grouping, we obtain:

Applying the Zero Product Property, we find two potential roots: $x=-\frac{1}{2}$ and $x=2$. However, we must recall that for any real angle $\theta$, the range of the sine function is restricted:

Because $x=2$ falls outside this interval, it is an extraneous solution. We discard it and proceed only with the valid condition:

On the unit circle, $\sin \theta =-\frac{1}{2}$ corresponds to angles where the $y$-coordinate is $-\frac{1}{2}$. This occurs in the third and fourth quadrants.

The reference angle for $\sin \alpha =\frac{1}{2}$ is $\alpha =\frac{\pi }{6}$. In the third quadrant, the principal angle is:

The general solution for $\sin \theta =\sin \alpha$ is given by $\theta =n\pi +(-1{)}^n\alpha$, where $n\in \mathbb{Z}$. Substituting our principal value $\alpha =\frac{7\pi }{6}$, we obtain the complete set of solutions:

This expression represents all possible angles satisfying the original equation. The solution is $\theta =n\pi +(-1{)}^n\frac{7\pi }{6}$ for all integers $n$.