If $0 \leq x<\frac{\pi}{2}$, then

$2 \sin x+\tan x \geq 3 x$

Consider the function

$f(x)=2 \sin x+\tan x-3 x$ for all $x \in[0, \pi / 2)$

We have,

$f'(x) =2 \cos x+\sec ^2 x-3$

$\Rightarrow f''(x)=-2 \sin x+2 \sec ^2 x \tan x$

$\Rightarrow f''(x)=-2 \sin x+\frac{2 \sin x}{\cos ^3 x}$

$\Rightarrow f''(x)=\frac{2 \sin x\left(1-\cos ^3 x\right)}{\cos ^3 x}>0$ for all $x \in(0, \pi / 2)$

⇒ f'(x) is increasing on $(0, \pi / 2)$

$\Rightarrow f'(x)>f'(0)$ for all $x \in(0, \pi / 2)$

$\Rightarrow f'(x)>0$ for all $x \in(0, \pi / 2)$

⇒ f(x) is increasing on $(0, \pi / 2)$

$\Rightarrow f(x)>f(0)$ for all $x \in(0, \pi / 2)$

$\Rightarrow 2 \sin x+\tan x-3 x>0$ for all $x \in(0, \pi / 2)$

$\Rightarrow 2 \sin x+\tan x>3 x$ for all $x \in(0, \pi / 2)$

At x = 0, we have

f(x) = 0

∴  $2 \sin x+\tan x \geq 3 x$ for all $x \in[0, \pi / 2)$