$$\sec{x}+\tan{x}=4$$ 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*}
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.
The other answer you found is probably $\cos x = 0$ , which doesn't fit because in the original equation $\tan x$ can't be evaluated that way. Remember $tan x = \frac{\sin x}{ \ cos x}$
Solving for sin$x $ you get:
sin$x=\frac {15}{17} $
Since both sin and cos values are positive, you can kick out the value in the 4th quadrant.
$$\tan x +\sec x =\frac{1+ \sin x}{ \cos x}$$ $$=\frac{(\cos\frac{x}{2}+ \sin \frac{x}{2})^2}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}$$ $$=\tan \left(\frac{\pi}{4}+\frac{x}{2} \right)$$
$$\frac{\pi}{4}+\frac{x}{2} =n\pi +tan^{-1}4$$ $$ x =2 n\pi +2 tan^{-1}4 -\frac{\pi}{2}$$ 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 .
HINT:
For finite non-zero $a,$
as $1=\sec^2x-\tan^2x=(\sec x+\tan x)(\sec x-\tan x)$
$$\sec x+\tan x=a\iff \sec x-\tan x=\dfrac1a$$
Add & subtract to $$\tan x,\sec x$$
Their signs will dictate the Quadrant of $x$
$$\sec{x}+\tan{x}=4$$ $$\sec{x}-4=\tan{x}$$ Squaring both sides, $$\sec^2 x-8\sec x+16=\sec^2 x-1$$ $$\sec{x}=\frac{17}{8}$$ $$\cos{x}=\frac{7}{18}$$ $$x=61.9°$$
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 $\sin x=4\cos x-1$.
$$\tan{x}=4-\sec{x}$$ Squaring both sides, $${\sec}^2{x}-1={\sec}^2{x}-8\sec{x}+16$$ $$8\sec{x}=17$$ $$\cos{x}=\frac{8}{17}$$
Solutions such that $\cos x=0$ are ruled out upfront as they do not belong to the domain.
Then you can rewrite the equation as
$$1+\sin x=4\cos x.$$
This is a classical linear form, which is easily transformed to
$$\cos(x+\phi)=c.$$
(See https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Linear_combinations.)
For $|c|<1$, the latter equation has exactly two solutions per period. But it turns out that one of them is $3\pi/2$ (which verifies $1+(-1)=4\cdot0$), precisely a forbidden value.
How about a geometric (projective) point of view ? The vector $\left( \begin{matrix} 1 + \sin x \\ \cos x \end{matrix} \right)$ should be non-zero and parallel to $\left( \begin{matrix} 4 \\ 1 \end{matrix} \right)$. The first vector traces a circle of radius 1 and centered at $(1,0)$ and should intersect the line of slope 1/4 passing through the origin. There are two intersections but the one at the origin is spurious in the present context.
In coordinates $(u,v)$ the equations to solve are $(u-1)^2+v^2=1$ and $u=4v$ yielding $17v^2 -8v=0$ or $v=\cos x= 8/17$ with the requirement $0<x<\pi/2$ (make a drawing) for the genuine solution.
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.
