1. **State the problem:** Solve the equation $$\sin^3 x (1 + \cot^2 x) = 1$$ for $$0 \leq x < 2\pi$$ using the unit circle. 2. **Recall the Pythagorean identity:** $$1 + \cot^2 x = \csc^2 x$$. 3. **Rewrite the equation using the identity:** $$\sin^3 x \cdot \csc^2 x = 1$$ 4. **Express $$\csc x$$ in terms of $$\sin x$$:** $$\csc x = \frac{1}{\sin x}$$, so $$\csc^2 x = \frac{1}{\sin^2 x}$$. 5. **Substitute into the equation:** $$\sin^3 x \cdot \frac{1}{\sin^2 x} = 1$$ 6. **Simplify the left side:** $$\sin^3 x \cdot \frac{1}{\sin^2 x} = \sin^{\cancel{3}} x \cdot \frac{1}{\sin^{\cancel{2}} x} = \sin x$$ 7. **So the equation reduces to:** $$\sin x = 1$$ 8. **Find all $$x$$ in $$[0, 2\pi)$$ where $$\sin x = 1$$:** From the unit circle, $$\sin x = 1$$ at $$x = \frac{\pi}{2}$$ only. **Final answer:** $$x = \frac{\pi}{2}$$ --- **Multiple Choice Question:** Identify the expression as a single trig identity: $$\frac{\sec^2 x - 1}{\sin^2 x}$$ Recall the Pythagorean identity: $$\sec^2 x - 1 = \tan^2 x$$ So, $$\frac{\sec^2 x - 1}{\sin^2 x} = \frac{\tan^2 x}{\sin^2 x}$$ Since $$\tan x = \frac{\sin x}{\cos x}$$, $$\tan^2 x = \frac{\sin^2 x}{\cos^2 x}$$ Substitute back: $$\frac{\tan^2 x}{\sin^2 x} = \frac{\frac{\sin^2 x}{\cos^2 x}}{\sin^2 x} = \frac{\sin^2 x}{\cos^2 x} \cdot \frac{1}{\sin^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$$ **Correct choice:** sec^2 x