Animated Solution for Mathematics - Trigonometry: Let $2{\sin }^{2}x+3\sin x-2>0$ and $x^2-x-2<0$ ($x$ is measured in radians). Then $x$ lies in the interval

First, let's tackle the trigonometric inequality:

At first glance, it looks like a complex trig problem, but it is a quadratic equation in disguise. By substituting $u=\sin x$, we transform this into:

Factoring this expression, we obtain:

Here is the key insight: we know that $\sin x$ is trapped between $-1$ and $1$. This means $u+2$ is always between $1$ and $3$, which is strictly positive.

We can safely divide by $(u+2)$ without changing the inequality sign. This leaves us with:

On the unit circle, this corresponds to the interval:

Now, let's pivot to the algebraic side:

This is a classic parabola opening upwards. Factoring the quadratic gives:

Using the wavy curve method (or sign scheme), we find the solution is the interval:

Now, we perform the final act: finding the intersection of the two solution sets. We have the intervals $(\frac{\pi }{6},\frac{5\pi }{6})$ and $(-1,2)$.

Using the approximation $\pi \approx 3.14$, we find:

Comparing these values, we see that the lower bound of the trig interval ($0.52$) is greater than the lower bound of the algebraic interval ($-1$). Furthermore, the upper bound of the algebraic interval ($2$) is less than the upper bound of the trig interval ($2.62$).

Thus, the intersection of these two conditions is:

$$x\in \left(\frac{\pi }{6},2\right)$$

You have mastered the art of dual constraints! Keep practicing, and remember that every complex problem is just a collection of simple, beautiful truths waiting to be uncovered.