1. **State the problem:** Solve the trigonometric equation $\sin^2 x - \cos^2 x - \sin x = 0$. 2. **Use the Pythagorean identity:** Recall that $\sin^2 x + \cos^2 x = 1$, so $\cos^2 x = 1 - \sin^2 x$. 3. **Substitute $\cos^2 x$ in the equation:** $$\sin^2 x - (1 - \sin^2 x) - \sin x = 0$$ 4. **Simplify the equation:** $$\sin^2 x - 1 + \sin^2 x - \sin x = 0$$ $$2\sin^2 x - \sin x - 1 = 0$$ 5. **Let $y = \sin x$ to form a quadratic:** $$2y^2 - y - 1 = 0$$ 6. **Solve the quadratic equation using the quadratic formula:** $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}$$ 7. **Calculate the roots:** - $$y_1 = \frac{1 + 3}{4} = 1$$ - $$y_2 = \frac{1 - 3}{4} = -\frac{1}{2}$$ 8. **Recall $y = \sin x$, so:** - $$\sin x = 1$$ - $$\sin x = -\frac{1}{2}$$ 9. **Find solutions for $\sin x = 1$:** $$x = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}$$ 10. **Find solutions for $\sin x = -\frac{1}{2}$:** $$x = \frac{7\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{11\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$ **Final answer:** $$x = \frac{\pi}{2} + 2k\pi, \quad x = \frac{7\pi}{6} + 2k\pi, \quad x = \frac{11\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$