Solving $\tan(x) = \cos(x)$ I've been trying to solve the following trigonometric equation unsuccessfully. My intuition was to reduce the degree of the equation from $4$ to $2$ so I could solve it as a quadratic equation, but my attempt so far have proved unsuccessful. Any help would be appreciated.
$$\begin{align}
\tan(x) &= \cos(x) \tag1\\[4pt]
\frac{\sin(x)}{\cos(x)} &=\cos(x) \tag2\\[4pt]
\frac{\sin^2(x)}{\cos^2(x)} &=\cos^2(x) \tag3\\[4pt]
\frac{1-\cos^2(x)}{\cos^2(x)} &=\cos^2(x) \tag4\\[4pt]
1-\cos^2(x) &=\cos^4(x) \tag5\\[6pt]
\cos^4(x) + \cos^2(x) - 1 &= 0  \tag6
\end{align}$$
It's at this point that I get stuck
 A: It is much simpler to deal with this as follows:\begin{align}\frac{\sin(x)}{\cos(x)}=\cos(x)&\iff\sin(x)=\cos^2(x)\\&\iff\sin(x)=1-\sin^2(x)\\&\iff\sin^2(x)+\sin(x)-1=0.\end{align}So, solve the equation $y^2+y-1=0$.
A: 
We could also construct the right triangle suggested by the equation $ \ \tan x \ = \ \cos x \ \ , $ where $ \ x \ $ is an angle (in the first quadrant).  We may set the length of the side opposite $ \ x \ $ equal to $ \ 1 \ \ , $ and write the ratios $ \ \tan x \ = \ \frac{1}{y} \ \ $ and $ \ \cos x \ = \ \frac{y}{y^2} \ = \ \frac{1}{y} \ \ . $  The "Pythagorean" Theorem then leads us to $ \ y^4 \ = \ y^2 + 1 \ \ . $
If we substitute $ \ u \ = \ y^2 \ \ $ to write this equation as $ \ u^2 \ = \ u \ + \ 1 \ \ , \ $ then divide through by $ \ u \ \ $ (which is positive), we have one of the "Golden Ratio" relations, $ \ u \ = \ 1 \ + \ \frac{1}{u} \ \ , $ with $ \ u \ = \ \phi \ = \ \frac{1 + \sqrt5}{2} \ \approx \ 1.61803 \ \ . $  We thus have the value in the first quadrant for our equation, $$   \tan x \ \ = \ \ \cos x \ \ = \ \ \frac{1}{\sqrt{\phi}} \ \ \approx \ \ 0.7862 \ \ . $$
There is a second result in the "principal circle", since  in the second quadrant we can have  $   \tan x \  =  \ \cos x \  = \  -\frac{1}{\sqrt{\phi}} \ \ . $  (There are no further solutions as $ \ \tan x \ $ and $ \ \cos x \ $ have opposite signs in the other two quadrants.)  The general solution to our equation is then
$$ x \ \ = \ \ \arccos \left(\frac{1}{\sqrt{\phi}} \right) \ + \ 2n\pi \ \ , \ \ \left[ \ \pi \ - \ \arccos \left( \frac{1}{\sqrt{\phi}} \right) \ \right] \ + \ 2n\pi \ \ . $$
[Interestingly, the first quadrant angle is extremely close to $ \ \frac23 \  : \  \arccos \left(\frac{1}{\sqrt{\phi}} \right) \ \approx \ 0.66624 \ \ . \ ] $
