1. **State the problem:** Prove the identity $$\frac{\sin x \tan x}{1-\cos x} \equiv 1 + \frac{1}{\cos x}$$ and then solve the equation $$\frac{\sin x \tan x}{1-\cos x} + 2 = 0$$ for $$0^\circ \leq x \leq 360^\circ$$. 2. **Recall formulas and identities:** - $$\tan x = \frac{\sin x}{\cos x}$$ - Important: $$1 - \cos x$$ is in the denominator, so $$\cos x \neq 1$$ to avoid division by zero. 3. **Prove the identity:** Start with the left side (LHS): $$\frac{\sin x \tan x}{1-\cos x} = \frac{\sin x \cdot \frac{\sin x}{\cos x}}{1-\cos x} = \frac{\sin^2 x}{\cos x (1-\cos x)}$$ Use the Pythagorean identity $$\sin^2 x = 1 - \cos^2 x$$: $$\frac{1 - \cos^2 x}{\cos x (1-\cos x)} = \frac{(1 - \cos x)(1 + \cos x)}{\cos x (1-\cos x)}$$ Cancel $$1 - \cos x$$ terms: $$\frac{1 + \cos x}{\cos x} = 1 + \frac{1}{\cos x}$$ This matches the right side (RHS), so the identity is proven. 4. **Solve the equation:** Given: $$\frac{\sin x \tan x}{1-\cos x} + 2 = 0$$ Using the proven identity: $$1 + \frac{1}{\cos x} + 2 = 0$$ Simplify: $$3 + \frac{1}{\cos x} = 0$$ $$\frac{1}{\cos x} = -3$$ $$\cos x = -\frac{1}{3}$$ 5. **Find solutions for $$x$$ in $$0^\circ \leq x \leq 360^\circ$$:** $$\cos x = -\frac{1}{3}$$ Cosine is negative in the second and third quadrants. Calculate reference angle: $$x_r = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.53^\circ$$ Solutions: - Second quadrant: $$180^\circ - 70.53^\circ = 109.47^\circ$$ - Third quadrant: $$180^\circ + 70.53^\circ = 250.53^\circ$$ 6. **Check domain restrictions:** $$\cos x \neq 1$$ is satisfied since $$\cos x = -\frac{1}{3}$$. **Final answers:** - Identity proven. - Solutions to the equation: $$x \approx 109.47^\circ, 250.53^\circ$$.