1. **State the problem:** Solve the inequality $$\frac{\sin x}{\tan x} + 1 > 0$$ for $x$. 2. **Recall the definitions and formulas:** - $\tan x = \frac{\sin x}{\cos x}$. - Substitute $\tan x$ in the expression: $$\frac{\sin x}{\tan x} = \frac{\sin x}{\frac{\sin x}{\cos x}} = \sin x \cdot \frac{\cos x}{\sin x}$$ 3. **Simplify the expression:** - Cancel $\sin x$ in numerator and denominator (noting $\sin x \neq 0$ to avoid division by zero): $$\frac{\sin x}{\tan x} = \cancel{\sin x} \cdot \frac{\cos x}{\cancel{\sin x}} = \cos x$$ 4. **Rewrite the inequality:** $$\cos x + 1 > 0$$ 5. **Solve the inequality:** - Subtract 1 from both sides: $$\cos x > -1$$ 6. **Analyze the inequality:** - Since $\cos x$ ranges between $-1$ and $1$, $\cos x > -1$ means $\cos x$ can be any value except exactly $-1$. 7. **Find values where $\cos x = -1$:** - $\cos x = -1$ at $x = \pi + 2k\pi$, where $k$ is any integer. 8. **Final solution:** - The inequality holds for all real $x$ except where $x = \pi + 2k\pi$. **Answer:** $$x \in \mathbb{R} \setminus \{\pi + 2k\pi : k \in \mathbb{Z}\}$$