Question:

Step 1: Let $t = \tan x$. The given equation can be rewritten as: $$2\sqrt{2}t^3 - \frac{54\sqrt{2}}{t^3} = 19$$ Step 2: Multiply the entire equation by $t^3$ to eliminate the denominator: $$2\sqrt{2}t^6 - 19t^3 - 54\sqrt{2} = 0$$ Step 3: Let $u = t^3$. Substituting $u$ into the equation yields a quadratic equation in terms of $u$: $$2\sqrt{2}u^2 - 19u - 54\sqrt{2} = 0$$ Step 4: Apply the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for $u$: $$u = \frac{19 \pm \sqrt{(-19)^2 - 4(2\sqrt{2})(-54\sqrt{2})}}{2(2\sqrt{2})}$$ $$u = \frac{19 \pm \sqrt{361 + 864}}{4\sqrt{2}}$$ $$u = \frac{19 \pm \sqrt{1225}}{4\sqrt{2}}$$ $$u = \frac{19 \pm 35}{4\sqrt{2}}$$ Step 5: Calculate the two possible values for $u$: $$u_1 = \frac{19 + 35}{4\sqrt{2}} = \frac{54}{4\sqrt{2}} = \frac{27}{2\sqrt{2}}$$ $$u_2 = \frac{19 - 35}{4\sqrt{2}} = \frac{-16}{4\sqrt{2}} = \frac{-4}{\sqrt{2}} = -2\sqrt{2}$$ Step 6: Consider the first case where $\tan^3\alpha = u_1$: $$\tan^3\alpha = \frac{27}{2\sqrt{2}}$$ Taking the cube root of both sides: $$\tan\alpha = \sqrt[3]{\frac{27}{2\sqrt{2}}} = \frac{3}{\sqrt[3]{2\sqrt{2}}} = \frac{3}{\sqrt{2}}$$ Step 7: Substitute this value of $\tan\alpha$ into the expression $(2\tan^2\alpha + \sqrt{2}\tan\alpha)$: $$2\left(\frac{3}{\sqrt{2}}\right)^2 + \sqrt{2}\left(\frac{3}{\sqrt{2}}\right) = 2\left(\frac{9}{2}\right) + 3 = 9 + 3 = 12$$ Step 8: Consider the second case where $\tan^3\alpha = u_2$: $$\tan^3\alpha = -2\sqrt{2}$$ Taking the cube root of both sides: $$\tan\alpha = \sqrt[3]{-2\sqrt{2}} = -\sqrt{2}$$ Step 9: Substitute this value of $\tan\alpha$ into the expression $(2\tan^2\alpha + \sqrt{2}\tan\alpha)$: $$2(-\sqrt{2})^2 + \sqrt{2}(-\sqrt{2}) = 2(2) - 2 = 4 - 2 = 2$$ Thus, the possible values of $(2\tan^2\alpha + \sqrt{2}\tan\alpha)$ are $12$ and $2$.