Solve the following equations: (1) \( 5 \cos ^{2} \theta+\cos \theta=0 \) (3) \( \sin x-6 \sin ^{2} x=0 \) (5) \( 8 \sin ^{2} \theta=2 ; \theta \in\left[0^{\circ} ; 720^{\circ}\right] \) (7) \( 6 \sin ^{2} \alpha-\sin \alpha=2 \) (9) \( 6=5 \cos \theta+6 \cos ^{2} \theta \)

To solve these equations, we can start with each one individually: (1) \( 5 \cos^{2} \theta + \cos \theta = 0 \) Factor out \(\cos \theta\): \(\cos \theta (5 \cos \theta + 1) = 0\). This gives us \(\cos \theta = 0\) or \(5 \cos \theta + 1 = 0\). From \(\cos \theta = 0\), \(\theta = 90^\circ, 270^\circ\). From \(5 \cos \theta + 1 = 0\), \(\cos \theta = -\frac{1}{5}\), giving \(\theta \approx 108.43^\circ, 251.57^\circ\). (3) \( \sin x - 6 \sin^{2} x = 0 \) Factor out \(\sin x\): \(\sin x (1 - 6 \sin x) = 0\). This gives us \(\sin x = 0\) (so \(x = 0, 180, 360, \ldots\)) or \(1 - 6 \sin x = 0\) solving for \(\sin x = \frac{1}{6}\), giving \(x \approx 9.59^\circ, 170.41^\circ\). (5) \( 8 \sin^{2} \theta = 2 \) Simplifying gives \(\sin^{2} \theta = \frac{2}{8} = \frac{1}{4}\). Thus, \(\sin \theta = \pm \frac{1}{2}\). For \(\sin \theta = \frac{1}{2}\), \(\theta = 30^\circ, 150^\circ\). For \(\sin \theta = -\frac{1}{2}\), \(\theta = 210^\circ, 330^\circ\). Now summing these gives us solutions in \([0^\circ, 720^\circ]\): \(30^\circ, 150^\circ, 210^\circ, 330^\circ, 390^\circ, 510^\circ, 570^\circ, 660^\circ\). (7) \( 6 \sin^{2} \alpha - \sin \alpha = 2 \) Rearranging, we can write the equation as \(6 \sin^{2} \alpha - \sin \alpha - 2 = 0\). Using the quadratic formula: \(\sin \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 48}}{12} = \frac{1 \pm 7}{12}\). This gives \(\sin \alpha = \frac{8}{12} = \frac{2}{3}\) or \(\sin \alpha = \frac{-6}{12} = -\frac{1}{2}\). Thus, \(\alpha \approx 41.81^\circ, 138.19^\circ, 210^\circ, 330^\circ\). (9) \( 6 = 5 \cos \theta + 6 \cos^{2} \theta \) Rearranging gives \(6 \cos^{2} \theta + 5 \cos \theta - 6 = 0\). Using the quadratic formula: \(\cos \theta = \frac{-5 \pm \sqrt{(5)^{2} - 4(6)(-6)}}{2(6)} = \frac{-5 \pm \sqrt{25 + 144}}{12} = \frac{-5 \pm 13}{12}\). Thus, \(\cos \theta = \frac{8}{12} = \frac{2}{3}\) or \(\cos \theta = \frac{-18}{12} = -\frac{3}{2}\) (not possible). So \(\cos \theta = \frac{2}{3}\) gives \(\theta \approx 48.19^\circ, 311.81^\circ\). By plugging these values in, we can work through the solutions for all these fun trigonometric puzzles!