1. **Problem:** Given the equation $$\frac{\sin^2 x}{\cos x} - \frac{\sin^2 x}{1+\cos x} = 1$$ for $$0^\circ \leq x \leq 360^\circ$$, find the value(s) of $$x$$. 2. **Formula and rules:** Use trigonometric identities and algebraic manipulation. Recall that $$\sin^2 x = 1 - \cos^2 x$$ and factor expressions carefully. 3. **Work:** Start with the equation: $$\frac{\sin^2 x}{\cos x} - \frac{\sin^2 x}{1+\cos x} = 1$$ Rewrite $$\sin^2 x$$ as $$1 - \cos^2 x$$: $$\frac{1 - \cos^2 x}{\cos x} - \frac{1 - \cos^2 x}{1+\cos x} = 1$$ Factor numerator: $$\frac{(1 - \cos x)(1 + \cos x)}{\cos x} - \frac{(1 - \cos x)(1 + \cos x)}{1+\cos x} = 1$$ Simplify second term: $$\frac{(1 - \cos x)(1 + \cos x)}{1+\cos x} = 1 - \cos x$$ So the equation becomes: $$\frac{(1 - \cos x)(1 + \cos x)}{\cos x} - (1 - \cos x) = 1$$ Multiply both sides by $$\cos x$$ to clear denominator: $$(1 - \cos x)(1 + \cos x) - (1 - \cos x) \cos x = \cos x$$ Expand: $$(1 - \cos^2 x) - (1 - \cos x) \cos x = \cos x$$ Distribute: $$1 - \cos^2 x - \cos x + \cos^2 x = \cos x$$ Simplify terms: $$1 - \cos x = \cos x$$ So: $$1 = 2 \cos x$$ $$\cos x = \frac{1}{2}$$ 4. **Solution:** For $$0^\circ \leq x \leq 360^\circ$$, $$\cos x = \frac{1}{2}$$ at: $$x = 60^\circ, 300^\circ$$ **Final answer:** $$x = 60^\circ, 300^\circ$$