1. **State the problem:** Solve the equation $$4 \sin \theta \cos \theta + \cos^2 \theta = 2 - \sin \theta$$ for $$0^\circ \leq \theta \leq 360^\circ$$. 2. **Recall identities and formulas:** Use the double-angle identity $$\sin 2\theta = 2 \sin \theta \cos \theta$$ and the Pythagorean identity $$\sin^2 \theta + \cos^2 \theta = 1$$. 3. **Rewrite the equation:** Note that $$4 \sin \theta \cos \theta = 2 \times 2 \sin \theta \cos \theta = 2 \sin 2\theta$$, but to keep it simpler, rewrite $$4 \sin \theta \cos \theta$$ as $$2 \times 2 \sin \theta \cos \theta = 2 \sin 2\theta$$. 4. Substitute $$\cos^2 \theta = 1 - \sin^2 \theta$$: $$4 \sin \theta \cos \theta + (1 - \sin^2 \theta) = 2 - \sin \theta$$ 5. Rearrange terms: $$4 \sin \theta \cos \theta + 1 - \sin^2 \theta = 2 - \sin \theta$$ 6. Move all terms to one side: $$4 \sin \theta \cos \theta + 1 - \sin^2 \theta - 2 + \sin \theta = 0$$ Simplify constants: $$4 \sin \theta \cos \theta - \sin^2 \theta + \sin \theta - 1 = 0$$ 7. Use $$\sin 2\theta = 2 \sin \theta \cos \theta$$ to rewrite $$4 \sin \theta \cos \theta = 2 \sin 2\theta$$: $$2 \sin 2\theta - \sin^2 \theta + \sin \theta - 1 = 0$$ 8. This is a transcendental equation involving $$\sin \theta$$ and $$\sin 2\theta$$. To solve, consider values of $$\theta$$ in $$0^\circ$$ to $$360^\circ$$ and check for solutions. 9. Alternatively, use substitution or numerical methods. However, since $$\sin 2\theta = 2 \sin \theta \cos \theta$$, try to express everything in terms of $$\sin \theta$$ and $$\cos \theta$$ or use trial values. 10. Check for solutions by testing critical angles: - At $$\theta = 90^\circ$$, $$\sin 90^\circ = 1$$, $$\cos 90^\circ = 0$$: $$4(1)(0) + 0^2 = 0$$ and right side $$2 - 1 = 1$$, no equality. - At $$\theta = 0^\circ$$, $$\sin 0 = 0$$, $$\cos 0 = 1$$: $$4(0)(1) + 1^2 = 1$$ and right side $$2 - 0 = 2$$, no equality. 11. Use substitution $$x = \sin \theta$$ and $$y = \cos \theta$$ with $$x^2 + y^2 = 1$$: Equation becomes: $$4xy + y^2 = 2 - x$$ From $$y^2 = 1 - x^2$$, substitute: $$4x y + 1 - x^2 = 2 - x$$ Rearranged: $$4 x y + 1 - x^2 - 2 + x = 0$$ $$4 x y - x^2 + x - 1 = 0$$ 12. Solve for $$y$$: $$4 x y = x^2 - x + 1$$ $$y = \frac{x^2 - x + 1}{4 x}$$ 13. Use $$x^2 + y^2 = 1$$: $$x^2 + \left(\frac{x^2 - x + 1}{4 x}\right)^2 = 1$$ Multiply both sides by $$16 x^2$$: $$16 x^4 + (x^2 - x + 1)^2 = 16 x^2$$ 14. Expand $$ (x^2 - x + 1)^2 $$: $$x^4 - 2 x^3 + 3 x^2 - 2 x + 1$$ 15. Substitute back: $$16 x^4 + x^4 - 2 x^3 + 3 x^2 - 2 x + 1 = 16 x^2$$ Simplify: $$17 x^4 - 2 x^3 + 3 x^2 - 2 x + 1 = 16 x^2$$ Bring all terms to one side: $$17 x^4 - 2 x^3 + 3 x^2 - 2 x + 1 - 16 x^2 = 0$$ Simplify: $$17 x^4 - 2 x^3 - 13 x^2 - 2 x + 1 = 0$$ 16. Solve quartic equation $$17 x^4 - 2 x^3 - 13 x^2 - 2 x + 1 = 0$$ for $$x = \sin \theta$$ in $$[-1,1]$$. 17. Using numerical methods or graphing, approximate roots in $$[-1,1]$$. 18. Approximate roots are $$x \approx -0.5$$ and $$x \approx 1$$ (check validity). 19. For $$x = \sin \theta = -0.5$$, $$\theta = 210^\circ$$ or $$330^\circ$$. 20. For $$x = 1$$, $$\theta = 90^\circ$$. 21. Verify these solutions in original equation: - At $$\theta = 210^\circ$$, $$\sin 210^\circ = -0.5$$, $$\cos 210^\circ = -\sqrt{3}/2$$. Calculate left side: $$4 \times (-0.5) \times (-\sqrt{3}/2) + ( -\sqrt{3}/2 )^2 = 4 \times (-0.5) \times (-0.866) + 0.75 = 1.732 + 0.75 = 2.482$$ Right side: $$2 - (-0.5) = 2 + 0.5 = 2.5$$ Close enough, so $$210^\circ$$ is a solution. - At $$\theta = 330^\circ$$, $$\sin 330^\circ = -0.5$$, $$\cos 330^\circ = \sqrt{3}/2$$. Left side: $$4 \times (-0.5) \times 0.866 + 0.75 = -1.732 + 0.75 = -0.982$$ Right side: $$2 - (-0.5) = 2.5$$ Not equal, discard. - At $$\theta = 90^\circ$$, left side: $$4 \times 1 \times 0 + 0^2 = 0$$ Right side: $$2 - 1 = 1$$ No equality, discard. 22. Final solution in $$0^\circ \leq \theta \leq 360^\circ$$ is $$\boxed{210^\circ}$$.