Show that 2sinx+tanx≥3x, where 0≤x< π/2 ​

Explanation

Let `f(x)=2 sinx + tan x-3x `
We have ` f'(x)=2 cos x + sec^(2)x -3 `
`=(2)/(sec x)+ sec^(2)x-3 `
`=(2+ sec^(3)x-3 sec x)/(sec x )`
`=(sec^(3)x-3sec x+2)/(sec x)`
`=((sec x-1)^(2)(sec+2))/(secx)` ........(i)
[The polynomial ` t^(3)-3t+2` factors as `(t-1)^(2)(t+2)`]
For ` 0 lt x lt (pi)/(2)`
`f'(x) gt 0` from (i)
Thus `f(x)` is an increasing function in `0 lt x lt (pi)/(2)`
`:. 2 sin x + tan x-3x gt 0, AA x in (0,(pi)/(2))` ........(A)
Note that , `f(0)=0`
i.e. ` 2 sin x + tan x -3 x =0 " for " x=0` .......(B)
Combining (A) and (B) , we have
` 2 sin x + tan x - 3x ge 0`
i.e., ` 2 sinx + tan x ge 3x AA x in [0,(pi)/(2))`