How to solve this trigonometric inequality: $k \sin\left(\frac {\pi x} {2}\right)\cos\left(\frac {\pi x} {2}\right)>0$? I was doing a physics problem today and I needed to find when some movement is accelerating and when it is decelerating. So after some calculations, I got the equation I needed. I know that it is correct because when plotted, I can see from graph that I got the expected solution. 
Unfortunately I don't know how to actually start solving the inequality. It looks like 
$$k \sin\left(\frac {\pi x} {2}\right)\cos\left(\frac {\pi x} {2}\right)>0$$ where $k$ is some huge sausage which is constant and positive.
 A: You can use the identity
$$\sin 2x = 2 \sin x \cos x$$
Thus you only need to determine when $$ \sin \pi x  > 0$$ and this happens if and only if $$ x \in (2m, 2m+1), m \in \mathbb{Z}$$
A: Since $k$ is positive, it is immaterial. You want to know the values of $x$ for which $\sin(\frac{\pi x}{2})$ and $\cos(\frac{\pi x}{2})$ have the same sign.
Each function has period $4$, so we can just look at what happens on $[0,4]$. $\sin(\frac{\pi x}{2})$ is positive when $0\lt\frac{\pi x}{2}\lt\pi$ (that is, when $0\lt x\lt 2$), and negative when $2\lt x \lt 4$. $\cos(\frac{\pi x}{2})$ is positive when $0\leq \frac{\pi x}{2}\lt \frac{\pi}{2}$ and when $\frac{3\pi}{2}\lt \frac{\pi x}{2}\lt 4$ (that is, when $0\leq x\lt 1$ and when $3\lt x\lt 4$). So the two have the same sign on $(0,1)$, and on $(2,3)$. Lather, rinse, and repeat. 
(Of course, this is easier to do using the identity that others have suggested; the approach of looking at the signs of the different factors is what you would use in general, or when there is no easy way of reducing the expression to a single item). 
A: Slighty simpler:$\;$ note $\rm\; ab > 0 \iff a/b > 0\;\;$ so $\;\;\sin(z)\cos(z)>0 \iff \tan(z) > 0$  
The rest is trivial using $\;\;\tan(-z) = -\tan(z)\;\;\:$ and $\:\;\;\tan(z+\pi) = \tan(z)$
A: $$ksin\left(\frac{\pi x}2\right)cos\left(\frac{\pi x}2\right)\gt0\\=>\frac k2sin(\pi x)\gt0$$ Now since k is constant and positive, just ignore k. $sin(\pi x)\gt0$ when $x\in (2x,2x+1)$ where $x \in Z$. 
