Solving a sine inequality, need intuition

Question

We have the function $f(x) = \sin (2x - \dfrac{1}{3} \pi)$ on the domain $ [0, 1\dfrac{1}{2} \pi ]$. Solve the inequality: $f(x) > \dfrac{1}{2}$

So I got to this point (I wrote it as in equality first):

$$x= \dfrac{1}{4} \pi + k\pi \vee x = \dfrac{7}{12} \pi + k\pi$$

This would yield the solutions $\dfrac{1}{4}\pi$, $1\dfrac{1}{4}\pi$ and $\dfrac{7}{12}\pi$. But my problem is, I don't know when $f(x) > 0.5$. I need some intuition (preferably using the unit circle) to figure out what the conditions would be.

Answer 1

Think of a point at angle $\theta$ on the unit circle. $\sin\theta$ is its $y$-coordinate. So, $\sin\theta$ is positive when $0 < \theta < 180$ degrees, because this puts our point in the upper half-plane (above the $x$-axis).

In your case, you need to put $\theta = 2x - \tfrac13\pi$, of course.

Does that help? If not ask again. I'm not really sure what you want.

Answer 2

$$\sin\left(2x-\frac{\pi}{3}\right)>\frac{1}{2}\;,\;\;x\in\left[0,\frac{3\pi}{2}\right]\iff \frac{\pi}{6}<2x-\frac{\pi}{3}<\frac{5\pi}{6}$$

Why? The easiest way to show the above is on the trigonometric (unit) circle: the sine function is positive only on the first and second quadrants, and it increases as the angle increases towards $\,\pi/2\,$ there...thus:

$$\frac{\pi}{6}<2x-\frac{\pi}{3}<\frac{5\pi}{6}\iff \frac{\pi}{2}<2x<\frac{7\pi}{6}\ldots$$

Answer 3

The most general and useful way to gain intuition about problems of this type is to sketch a graph of the function.