If we have an expression as ${{\log }_{\tan x}}\left( 2+4{{\cos }^{2}}x \right)=2$ then $x=$ \[\]
A.$n\pi \pm \dfrac{\pi }{6},n\in Z$\[\]
B. $2n\pi \pm \dfrac{\pi }{3},n\in Z$\[\]
C. $n\pi \pm \dfrac{\pi }{4},n\in Z$\[\]
D. $\left( 2n+1 \right)\pm \dfrac{\pi }{2},n\in Z$\[\]

Hint: We use the definition of logarithm and convert the equation to $2+4{{\cos }^{2}}x={{\tan }^{2}}x$. We convert the tangent and sine into cosines $\tan \theta =\dfrac{\sin \theta }{\cos \theta },{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $. We put ${{\cos }^{2}}x=t$ and solve the quadratic equation for $t$. We finally use the theorem that the solutions of $\cos x=\cos \alpha $ are $x=2n\pi \pm \alpha $where $\alpha $ is principal value and $n$ is arbitrary integer.

Complete step-by-step solution
We know that the logarithm is the inverse function of the exponentiation function. That means the logarithm of a given number $x$ is the exponent to which another fixed number, the base $b$ must be raised, to produce that number $x$, which means if ${{b}^{y}}=x$ then the logarithm denoted as log and calculated as
\[{{\log }_{b}}x=y\]
We take the converse to have,
\[{{\log }_{b}}x=y\Leftrightarrow {{b}^{y}}=x\]
We are given in the question an equation involving logarithmic and trigonometric function as follows.
\[{{\log }_{\tan x}}\left( 2+4{{\cos }^{2}}x \right)=2\]
We us definition of logarithm to have
\[2+4{{\cos }^{2}}x={{\left( \tan x \right)}^{2}}={{\tan }^{2}}x\]
We use the formula to convert tangent of an angle to sine and cosine $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ in the above step and have,
\[\Rightarrow 2+4{{\cos }^{2}}x=\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}\]
We multiply ${{\cos }^{2}}x$ on both sides of the equation in the above step to have.
\[\Rightarrow 2{{\cos }^{2}}x+4{{\cos }^{4}}x={{\sin }^{2}}x\]
We use the Pythagorean identity ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ to convert the sine into cosine in the above step to have,
\[\begin{align}
  & \Rightarrow 2{{\cos }^{2}}x+4{{\cos }^{4}}x=1-{{\cos }^{2}}x \\
 & \Rightarrow 4{{\cos }^{4}}x+3{{\cos }^{2}}x-1=0 \\
 & \Rightarrow 4{{\left( {{\cos }^{2}}x \right)}^{2}}+3{{\cos }^{2}}x-1=0 \\
\end{align}\]
We put ${{\cos }^{2}}x=t$ in the above equation to have a quadratic equation in $t$,
\[\Rightarrow 4{{t}^{2}}+3t-1=0\]
We solve the quadratic equation by splitting the middle term $3t$ to $4t-t$ because $4t\left( -t \right)=-4{{t}^{2}}$. So we have
\[\begin{align}
  & \Rightarrow 4{{t}^{2}}+4t-t-1=0 \\
 & \Rightarrow 4t\left( t+1 \right)-\left( t+1 \right)=0 \\
 & \Rightarrow \left( 4t-1 \right)\left( t+1 \right)=0 \\
 & \Rightarrow 4t-1=0,t+1=0 \\
 & \Rightarrow t=\dfrac{1}{4}\text{ or }t=-1 \\
\end{align}\]
The above step gives us the two roots of the quadratic equation $t=\dfrac{1}{4},-1$. We put back $t={{\cos }^{2}}x$ to have,
\[\Rightarrow {{\cos }^{2}}x=\dfrac{1}{4}\text{ or }{{\cos }^{2}}x=-1\]
We reject the possibility that ${{\cos }^{2}}x=-1$ because ${{\cos }^{2}}x$ is always a positive quantity and proceed with other equation by factoring using the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
\[\begin{align}
  & \Rightarrow {{\cos }^{2}}x=\dfrac{1}{4}\text{ } \\
 & \Rightarrow \text{co}{{\text{s}}^{2}}x-{{\left( \dfrac{1}{2} \right)}^{2}}=0 \\
 & \Rightarrow \left( \cos x+\dfrac{1}{2} \right)\left( \cos -\dfrac{1}{2} \right)=0 \\
 & \Rightarrow \cos x=\dfrac{-1}{2}\text{ or }\cos x=\dfrac{1}{2}\text{ } \\
\end{align}\]
We know that the solutions of $\cos x=\cos \alpha $ are $x=2n\pi \pm \alpha $ where $\alpha $ is principal value and $n$ integer. We know that the principal angle for $\dfrac{1}{2},\dfrac{-1}{2}$ are $\dfrac{\pi }{3},\dfrac{2\pi }{3}$ since we have $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2},\cos \left( \dfrac{2\pi }{3} \right)=\dfrac{-1}{2}$. We use it n the next step,
\[\begin{align}
  & \Rightarrow \cos x=\cos \left( \dfrac{2\pi }{3} \right)\text{ or }\cos x=\cos \left( \dfrac{\pi }{3} \right)\text{ } \\
 & \Rightarrow x=2n\pi \pm \dfrac{2\pi }{3}\text{ or }x=2n\pi \pm \dfrac{\pi }{3} \\
\end{align}\]
We find all the solutions taking union of the solutions and have
\[\therefore x=n\pi \pm \dfrac{\pi }{3}\text{ },n\in Z\]
So the correct choice is B.

Note: We note that tangent function is not defined for $x=\left( 2n+1 \right)\dfrac{\pi }{2}$ which is the obtained solution satisfies and logarithmic function only takes positive value which $2+4{{\cos }^{2}}x$ satisfies for all obtained values of $x$. We must not confuse between solutions of $\cos x=\cos \alpha $ and $\cos x=0$ whose solutions are $x=\left( 2n+1 \right)\dfrac{\pi }{2}$. If $\cos \alpha =k$ such that $\alpha $ is the first value cosine takes for $k$ before repeating then $\alpha $ is called principal value.