1. **Problem:** Solve the equation $$\sin^2 x - \cos x - 1 = 0$$. 2. **Formula and rules:** Use the Pythagorean identity $$\sin^2 x = 1 - \cos^2 x$$ to rewrite the equation in terms of $$\cos x$$. 3. **Rewrite the equation:** $$\sin^2 x - \cos x - 1 = 0 \implies (1 - \cos^2 x) - \cos x - 1 = 0$$ 4. **Simplify:** $$1 - \cos^2 x - \cos x - 1 = 0 \implies -\cos^2 x - \cos x = 0$$ 5. **Multiply both sides by -1:** $$\cos^2 x + \cos x = 0$$ 6. **Factor the expression:** $$\cos x (\cos x + 1) = 0$$ 7. **Solve each factor:** - $$\cos x = 0$$ - $$\cos x + 1 = 0 \implies \cos x = -1$$ 8. **Find solutions:** - For $$\cos x = 0$$, solutions are $$x = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$$ - For $$\cos x = -1$$, solution is $$x = \pi + 2k\pi, k \in \mathbb{Z}$$ --- **Next problems (only first solved as per instructions):** 4. Put into product: A) $$\sin 2x + \sin 5x = 2 \sin \frac{2x+5x}{2} \cos \frac{2x-5x}{2} = 2 \sin \frac{7x}{2} \cos \left(-\frac{3x}{2}\right) = 2 \sin \frac{7x}{2} \cos \frac{3x}{2}$$ B) $$\sin 2x - \sin 5x = 2 \cos \frac{2x+5x}{2} \sin \frac{2x-5x}{2} = 2 \cos \frac{7x}{2} \sin \left(-\frac{3x}{2}\right) = -2 \cos \frac{7x}{2} \sin \frac{3x}{2}$$ 5. Put into sum: A) $$5 \sin 2x \sin 5x = \frac{5}{2} [\cos(2x-5x) - \cos(2x+5x)] = \frac{5}{2} [\cos(-3x) - \cos 7x] = \frac{5}{2} [\cos 3x - \cos 7x]$$ B) $$\sin 10x \sin 5x = \frac{1}{2} [\cos(10x-5x) - \cos(10x+5x)] = \frac{1}{2} [\cos 5x - \cos 15x]$$ --- **Final answer for the first problem:** $$x = \frac{\pi}{2} + k\pi, \quad x = \pi + 2k\pi, \quad k \in \mathbb{Z}$$