Domain of Trigonometric Inequality

Question

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?

Answer 1

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

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

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

while

2. $\sin(x)>\frac{1}{2}$ 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 $\frac{3\pi}{2}\sim 4,71$ )