1. The problem asks to solve the equation $$\frac{2 \sin x - \cos x}{2 \cos x + \sin x} = 3$$ for $x$. 2. We start by multiplying both sides by the denominator to clear the fraction: $$2 \sin x - \cos x = 3(2 \cos x + \sin x)$$ 3. Expanding the right side: $$2 \sin x - \cos x = 6 \cos x + 3 \sin x$$ 4. Rearranging terms to isolate sine and cosine on one side: $$2 \sin x - 3 \sin x = 6 \cos x + \cos x$$ 5. Simplify both sides: $$-\sin x = 7 \cos x$$ 6. Divide both sides by $\cos x$ (assuming $\cos x \neq 0$): $$\frac{-\sin x}{\cos x} = 7$$ 7. Using the identity $\tan x = \frac{\sin x}{\cos x}$, we get: $$-\tan x = 7 \implies \tan x = -7$$ 8. Therefore, the solution is all $x$ such that: $$\tan x = -7$$ 9. From the options given, the value of the expression is $-7$, so the correct answer is D) -7.