Domain of Trigonometric Inequality The problem: Find values of $x$ on the interval $[0,\pi]$ for which $\cos(x)< \sin(2x)$
My answer came out to be $0.52< x<1.57$ or $2.62< x\leq π$. However, the textbook's answer is slightly different: $0.52< x<1.57$ or $2.62< x<π$. At $x=π$, $\cos(x)< \sin(2x)$, and $x$ can be $[0,π]$ so why is $π$ left out of the solution? 

$\cos(x)-2\sin(x)\cos(x)<0$ $\leftrightarrow$
$\cos(x)[1-2\sin(x)]<0$ 


*

*$\cos(x)<0$ if and only if $ \frac{\pi}{2}+2k\pi<x<\frac{3\pi}{2}+2k\pi$


while 


*$\sin(x)>\frac{1}{2}$ if and only if 
$\frac{\pi}{6}+2k\pi<x<\frac{5\pi}{6}+2k\pi$ 


So the solution is 
$\frac{\pi}{6}+2k\pi<x<\frac{\pi}{2}+2k\pi$
or 
$\frac{5\pi}{6}+2k\pi<x<\frac{3\pi}{2}+2k\pi$
So $\pi$ is oviously a  solution of your problem 
(You can observe that $\frac{\pi}{6}\sim 0,52$ , $\frac{\pi}{2}\sim 1,57$, $\frac{5\pi}{6}\sim 2,62 $ and $\frac{3\pi}{2}\sim 4,71$ )
