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.

Multiplying through by $\cos(x)$ and re-writing in terms of sine:
$$1 + \sin(x) = 4\cos(x) \implies 1 + \sin(x) = 4\sqrt{1 - \sin^2(x)}$$
And now we square both sides using the latter representation:
$$\sin^2(x) + 2\sin(x) + 1 = 16 - 16\sin^2(x)$$
From here, I write $y = \sin(x)$ for ease of notation, and rearrange:
$$y^2 + 2y + 1 = 16 - 16y^2 \implies 17y^2 + 2y - 15 = 0 \implies (17y - 15)(y + 1) = 0$$
The possible solutions now arise from checking $\sin(x) = 15/17$ and $\sin(x) = -1$.
Note that checking is, in this case, necessary: Early on we cleared denominators of $\cos(x)$ (so a final scenario in which $\cos(x) = 0$ will not yield a solution) and later on we squared, which can also introduce new solutions. 
For the two possible solutions that we found, the former value satisfies the initial equation; the latter value does not.
