If 2 - cos^2 θ = 3 sin θ cos θ, sin θ ≠ cos θ then tan θ is

$$2-cos^2 \theta = 1+1-cos^2 \theta = 1+sin^2 \theta$$
Dividing the LHS and RHS by $$cos^2\theta$$
$$1+sin^2 \theta = sec^2 \theta + tan^2 \theta$$
$$3sin \theta cos \theta = 3 tan \theta$$
$$sec ^2 \theta + tan^2 \theta = 3tan \theta$$
$$sec ^2 \theta = 1+tan^2 \theta$$
$$1+tan^2 \theta+tan^2 \theta = 3tan \theta$$
$$1+2tan^2 \theta = 3tan \theta$$
$$2tan^2 \theta - 3tan \theta + 1 =0$$
let x=$$tan \theta$$
The equation becomes
$$2x^2 + 3x + 1=0$$
On solving for x we get x=$$1$$ and x=$$\frac{1}{2}$$
Option A is the correct answer.