If A = x ∈ [ 0 , 2 π ] / tan x - tan 2 x > 0 and B = x ∈ [ 0 , 2 π ] / | sin x |…

We have

$A=\left\{x\in [0,2\pi ]/\tan x-{\tan }^{2}x>0\right\}$

$B=\left\{x\in [0,2\pi ]/\left|\sin x\right|<\frac{1}{2}\right\}$

For set $A$

$\tan x-{\tan }^{2}x>0$

$\Rightarrow \tan x\left(1-\tan x\right)>0$

$\Rightarrow 0<\tan x<1$

  From the graph, it is clear that for$x\in \left(0,\frac{\mathrm{\pi }}{4}\right)\cup \left(\mathrm{\pi },\frac{5\mathrm{\pi }}{4}\right)\Rightarrow \tan x\in \left(0,1\right)$, 

For set $B$

$\left|\sin x\right|<\frac{1}{2}$

From the graph, it is clear that, for $x\in \left(0,\frac{\mathrm{\pi }}{6}\right)\cup \left(\frac{5\mathrm{\pi }}{6},\frac{7\mathrm{\pi }}{6}\right)\cup \left(\frac{11\mathrm{\pi }}{6},2\mathrm{\pi }\right)\Rightarrow \left|\sin x\right|<\frac{1}{2}$

So, common solution is

$\left(0,\frac{\pi }{6}\right)\cup \left(\pi ,\frac{7\pi }{6}\right)$