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.

Using $t$-formula
Let $\displaystyle t=\tan \frac{x}{2}$, then $\displaystyle \cos x=\frac{1-t^2}{1+t^2}$ and $\displaystyle \tan x=\frac{2t}{1-t^2}$.
Now
\begin{align*}
  \frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2} &=4 \\
  \frac{(1+t)^{2}}{1-t^2} &= 4 \\
  \frac{1+t}{1-t} &= 4 \quad \quad (t\neq -1) \\
  t &= \frac{3}{5} \\
  \tan \frac{x}{2} &= \frac{3}{5} \\
  x &=2\left( n\pi +\tan^{-1} \frac{3}{5} \right) \\
  x &= 2\tan^{-1} \frac{3}{5} \quad \quad (0<x<2\pi)
\end{align*}
