Solve $\sec (x) + \tan (x) = 4$ $$\sec{x}+\tan{x}=4$$
Find $x$ for $0<x<2\pi$.
Eventually I get $$\cos x=\frac{8}{17}$$
$$x=61.9^{\circ}$$
The answer I obtained is the only answer, another respective value of $x$ in $4$-th quadrant does not solve the equation, how does this happen?  I have been facing the same problem every time I solved this kind of trigonometric equation.

$$\tan x +\sec x =\frac{1+ \sin x}{ \cos x}$$
$$=\frac{(\cos\frac{x}{2}+ \sin \frac{x}{2})^2}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}$$
$$=\tan \left(\frac{\pi}{4}+\frac{x}{2} \right)$$
$$\frac{\pi}{4}+\frac{x}{2} =n\pi +tan^{-1}4$$
$$ x =2 n\pi +2 tan^{-1}4 -\frac{\pi}{2}$$
For solution to be in $[0,2\pi]$ 
$$ n =0 $$
$x=1.080 $ or $ 61.9^{\circ}$
You can also see from the graph that in $[0,2\pi]$ there is only solution lying .

