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.

Rewrite your equation as
$$
\frac{1}{\cos x}+\frac{\sin x}{\cos x}=4
$$
that becomes
$$
\frac{1+\sin x}{\cos x}=4
$$
This reminds the formula for the tangent of the half angle
$$
\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha}
$$
but sine and cosine are mixed up. Not a problem: set $x=\pi/2-t$ and take the reciprocal:
$$
\frac{\sin t}{1+\cos t}=\frac{1}{4}
$$
that's
$$
\tan\frac{t}{2}=\frac{1}{4}
$$
Thus
$$
t=2\arctan\frac{1}{4}
$$
so
$$
x=\frac{\pi}{2}-2\arctan\frac{1}{4}\approx1.08084
$$
In degrees, $x=61.92757^\circ$.

A different strategy could be rewriting the equation as
$$
\sin x=4\cos x-1
$$
and squaring (but this may add spurious solutions):
$$
1-\cos^2x=16\cos^2x-8\cos x+1
$$
that gives
$$
\cos x(17\cos x-8)=0
$$
We know that $\cos x\ne0$ (because the original problem has $\sec x$) and so we remain with $\cos x=8/17$.
This leads to $x=\arccos(8/17)$ or $x=2\pi-\arccos(8/17)$, but the second solution must be discarded, because it leads to $\sin x<0$, whereas
$$
4\cdot\frac{8}{17}-1>0
$$
so this is incompatible with $\sin x=4\cos x-1$.
