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 ?