1. **State the problem:** We need to show that the equation $$2 \tan^2 \theta \sin^2 \theta = 1$$ can be rewritten as $$2 \sin^4 \theta + \sin^2 \theta - 1 = 0$$. 2. **Recall the identity:** $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$, so $$\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$$. 3. **Substitute into the original equation:** $$2 \tan^2 \theta \sin^2 \theta = 2 \cdot \frac{\sin^2 \theta}{\cos^2 \theta} \cdot \sin^2 \theta = 2 \frac{\sin^4 \theta}{\cos^2 \theta} = 1$$ 4. **Multiply both sides by $$\cos^2 \theta$$:** $$2 \sin^4 \theta = \cos^2 \theta$$ 5. **Use the Pythagorean identity:** $$\cos^2 \theta = 1 - \sin^2 \theta$$, so $$2 \sin^4 \theta = 1 - \sin^2 \theta$$ 6. **Rearrange to standard polynomial form:** $$2 \sin^4 \theta + \sin^2 \theta - 1 = 0$$ --- 7. **Solve the equation:** Let $$x = \sin^2 \theta$$, then $$2x^2 + x - 1 = 0$$ 8. **Use the quadratic formula:** $$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$ 9. **Calculate roots:** - $$x = \frac{-1 + 3}{4} = \frac{2}{4} = 0.5$$ - $$x = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$$ (discard since $$\sin^2 \theta \geq 0$$) 10. **So, $$\sin^2 \theta = 0.5$$, which means $$\sin \theta = \pm \frac{\sqrt{2}}{2}$$. 11. **Find $$\theta$$ in $$0^\circ \leq \theta \leq 360^\circ$$:** - $$\sin \theta = \frac{\sqrt{2}}{2}$$ at $$\theta = 45^\circ, 135^\circ$$ - $$\sin \theta = -\frac{\sqrt{2}}{2}$$ at $$\theta = 225^\circ, 315^\circ$$ **Final answer:** $$\theta = 45^\circ, 135^\circ, 225^\circ, 315^\circ$$