$\text{ If }\sin x+\cos x+\tan x+cosecx+\sec x+\cot x=7, \text{ then }\sin 2x=$

$\left(\sin x+\cos x\right)+\frac{1}{\sin x\cos x}+\frac{\sin x+\cos x}{\sin x\cos x}=7$

$\Rightarrow \left(\sin x+\cos x\right)\left(1+\frac{1}{\sin x\cos x}\right)=7-\frac{1}{\sin x\cos x}$

Squaring on both sides we get

$\begin{array}{l}\Rightarrow \sin 2x\left({\sin }^{2}2x-44\sin 2x+36\right)=0\\ \sin 2x\ne 0\\ \Rightarrow \sin 2x=22-8\sqrt{7}\end{array}$