$ \text{sec}^2x + 3\text{cosec}^2x =8 $ Solving this trigonometric equation. I have found two different methods of solving this trigonometric equation :
$$ \text{sec}^2x + 3\text{cosec}^2x =8 
$$
But these methods give different answers.
Solution 1
$$ \text{sec}^2x + 3\text{cosec}^2x =8 $$
$$\implies \frac{1}{\text{cos}^2x} + \frac{3}{\text{sin}^2x} =8 
$$
$$\implies 3{\text{cos}^2x} + {\text{sin}^2x} =8(\text{cos}^2x ×\text{sin}^2x)$$
$$\implies 2{\text{cos}^2x} + 1 =8(\text{cos}^2x ×(1-\text{cos}^2x))$$
$$\implies 8{\text{cos}^4x}  -6\text{cos}^2x +1 =0$$
$$\implies (2\text{cos}^2x-1)(4\text{cos}^2x-1)=0
$$
$$ \implies x= 2nπ±\frac{π}{4},2nπ±(π-\frac{π}{4}),2nπ±\frac{π}{3},2nπ±(π-\frac{π}{3}),
$$
Solution 2
$$ \text{sec}^2x + 3\text{cosec}^2x =8 $$
$$ \implies 1+\text{tan}^2x + 3 + 3\text{cot}^2x =8 $$
$$\implies {\text{tan}^2x} + \frac{3}{\text{tan}^2x} =4 
$$
$$ \implies \text{tan}^4x -4\text{tan}^2x + 3 =0 $$
$$ \implies (\text{tan}^2x -1)(\text{tan}^2x-3) =0 $$
$$ \implies x=nπ±\frac{π}{3},nπ±\frac{π}{4} $$
Now my question is that which one should I reject and why ?
 A: Using $\cos^{2}x + \sin^{2}x = 1$, from your first solution method and first factor, $2\cos^{2}x - 1 = 0 \; \to \; \cos^{2}x = \frac{1}{2}$, so $\sin^{2}x = 1 - \frac{1}{2} = \frac{1}{2}$ and, thus, $\tan^{2}x = \frac{1/2}{1/2} = 1$. The second factor gives $4\cos^{2}x - 1 = 0 \; \to \; \cos^{2}x = \frac{1}{4}$, so $\sin^{2}x = 1 - \frac{1}{4} = \frac{3}{4}$, which means $\tan^{2}x = \frac{3/4}{1/4} = 3$. Note these match the $\tan^{2}x$ values obtained from your second solution method's factorization.
Your first two sets of values of $x= 2n\pi\pm\frac{\pi}{4},2n\pi\pm(\pi-\frac{\pi}{4})$ can be combined into just $n\pi\pm\frac{\pi}{4}$ (since the $+$ part of $2n\pi\pm(\pi-\frac{\pi}{4})$ is $(2n+1)\pi-\frac{\pi}{4}$, i.e., all odd values of $n$ with $n\pi-\frac{\pi}{4}$, with the $-$ part being $(2n-1)\pi+\frac{\pi}{4}$, i.e., all odd values of $n$ with $n\pi+\frac{\pi}{4}$), with this just being the solutions to $\cos^{2}x = \frac{1}{2} \; \to \; \cos x = \pm\frac{1}{\sqrt{2}}$. For the second set of solutions, $\cos^{2}x = \frac{1}{4} \; \to \; \cos x = \pm\frac{1}{2}$, which has solutions of $n\pi\pm\frac{\pi}{3}$ (your second set of values have the same basic issue as with the first set, i.e., they can be combined into just the one set of values I give here, and which you also gave in your second solution).
Thus, the sets of $x$ values from each of your $2$ solution techniques actually match each other.
