# Solving a sine inequality, need intuition

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.

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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.

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This really helps, I figured it out. The useful insight you gave me is that we need to view $\theta = 2x - \dfrac{1}{3} \pi$ – Ylyk Coitus Mar 10 '13 at 13:07

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

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:

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

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I have the same answer now, but my textbook also adds $1\dfrac{1}{4}\pi < x \leq 1\dfrac{1}{2} \pi$, is this incorrect, it seems to me it is (using the unit circle) – Ylyk Coitus Mar 10 '13 at 13:12
That makes no sense, @YlykCoitus , as sine is negative in the third quadrant...perhaps you forgot to add some absolute value, or may something's squared somewhere...? – DonAntonio Mar 10 '13 at 13:15
Nope, I didn't. The book is just wrong, as always. They were double wrong actually. I added the $\leq$ sign in the answer, they have it as $1 \dfrac{1}{4} \pi < x < 1\dfrac{1}{2} \pi$ , which is even more wrong right? – Ylyk Coitus Mar 10 '13 at 13:19
Well...hehe...wrong, wronger...It is wrong, period. – DonAntonio Mar 10 '13 at 13:21
BTW, what book is that? When I was an undergraduate I used to "hunt" for mistakes, misprints, etc. in mathematics books. That could be fun... – DonAntonio Mar 10 '13 at 13:22

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

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Of course, that is also true, thanks. I however particularly wanted to use the unit circle in my understanding of this – Ylyk Coitus Mar 10 '13 at 19:56