Determine the general solution of tan x = 2 sin 2x where cos x < 0.

Recall that $\tan x = \frac{\sin x}{\cos x}$ and $\sin 2x = 2\sin x \cos x$.
Substitute these into the given equation:

Multiply both sides by $\cos x$ (note: $\cos x \neq 0$ for $\tan x$ to be defined, so we must check solutions where $\cos x = 0$ separately if they arise from the original equation, but here we are given $\cos x < 0$, so $\cos x \neq 0$ is already implied).

Move all terms to one side:

Factor out $\sin x$:

This gives two separate cases:
Case 1: $\sin x = 0$
Case 2: $1 - 4\cos^2 x = 0$

For $\sin x = 0$, the general solution is $x = 180^\circ k$, where $k \in \mathbb{Z}$.
Now, we apply the condition $\cos x < 0$.

• If $k$ is an even integer (e.g., $k=0, 2, -2, \dots$), then $x = 0^\circ, 360^\circ, -360^\circ, \dots$. For these values, $\cos x = 1$, which does not satisfy $\cos x < 0$.
• If $k$ is an odd integer (e.g., $k=1, 3, -1, \dots$), then $x = 180^\circ, 540^\circ, -180^\circ, \dots$. For these values, $\cos x = -1$, which satisfies $\cos x < 0$. So, the solutions from Case 1 that satisfy $\cos x < 0$ are $x = 180^\circ + 360^\circ k$, where $k \in \mathbb{Z}$.

For $1 - 4\cos^2 x = 0$:

Take the square root of both sides:

Now, apply the condition $\cos x < 0$. This means we only consider $\cos x = -\frac{1}{2}$.
For $\cos x = -\frac{1}{2}$:
The reference angle is $60^\circ$. Since $\cos x$ is negative, $x$ lies in the second or third quadrants.

• In the second quadrant: $x = 180^\circ - 60^\circ = 120^\circ$.
• In the third quadrant: $x = 180^\circ + 60^\circ = 240^\circ$. The general solutions for these are: $x = 120^\circ + 360^\circ k$, where $k \in \mathbb{Z}$. $x = 240^\circ + 360^\circ k$, where $k \in \mathbb{Z}$.

The general solutions satisfying $\cos x < 0$ are:

where $k \in \mathbb{Z}$.

The general solution is $\boxed{x = 180^\circ + 360^\circ k, x = 120^\circ + 360^\circ k, x = 240^\circ + 360^\circ k, k \in \mathbb{Z}}$.