Animated Solution for Mathematics - Trigonometry: The values of $\theta \in (0,2\pi )$ for which $2{\sin }^2\theta -5\sin \theta +2>0$, are

The given expression is $2{\sin }^2\theta -5\sin \theta +2>0$. To simplify our perspective, we replace the trigonometric term $\sin \theta$ with an algebraic variable $x$.

This transforms the problem into the quadratic inequality:

We factor the quadratic by splitting the middle term. We seek two numbers that multiply to $4$ (the product of $2$ and $2$) and add to $-5$. These numbers are $-4$ and $-1$.

Rewriting the expression, we have:

Grouping the terms gives:

We must remember that $x=\sin \theta$. The fundamental property of the sine function is that it is bounded:

Consider the second factor, $(x-2)$. Since the maximum value of $x$ is $1$, the maximum value of $(x-2)$ is $1-2=-1$.

This implies that $(x-2)$ is strictly negative for all values of $\theta$.

For the product $(2x-1)(x-2)$ to be positive, and knowing $(x-2)$ is always negative, the first term $(2x-1)$ must also be negative. A negative multiplied by a negative yields a positive result.

Therefore, we require:

Substituting back for $\sin \theta$:

We now find where the sine curve lies below the horizontal line $y=1/2$ within the domain $(0,2\pi )$. We know that $\sin \theta =1/2$ at $\theta =\pi /6$ and $\theta =5\pi /6$.

By analyzing the sine wave, the curve is below the line $y=1/2$ in the intervals $(0,\pi /6)$ and $(5\pi /6,2\pi )$.

The final solution set is: