<s>[INST] 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.
 [/INST] If $\sec x+ \tan x=4 \\
\implies \frac{1+\sin x}{\cos x}=4\\
\implies 1+\sin x=4\cos x$
That does not help! So let's square it both sides
$\implies 1+\sin^2x+2\sin x=16\cos^2x=16-16\sin^2x
\implies 17\sin^2x+2\sin x-15=0$
Assuming, $\sin x=y$, makes the above equation quadratic in y having 2 solutions. So,
$17y^2+17y-15y-15=0\\
\implies 17y(y+1)-15(y+1)=0\\
\implies (17y-15)(y+1)=0
$
So, $y=\frac{15}{17}$ and $y=-1$
So either $x =270^\circ$ or $x\approx 62^{\circ}$
EDIT: I looked at other answers and was wondering what solid reasoning is behind not accepting  $x =270^\circ$. 
This below.

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