1. **State the problem:** Solve the equation $$(\tan 3\theta + \sec 3\theta)^2 = 6$$ for $$0^\circ \leq \theta \leq 180^\circ$$. 2. **Recall the identity:** We know that $$\tan x + \sec x = \frac{\sin x}{\cos x} + \frac{1}{\cos x} = \frac{\sin x + 1}{\cos x}$$. 3. **Rewrite the equation:** Let $$x = 3\theta$$, then the equation becomes: $$\left(\tan x + \sec x\right)^2 = 6$$ which is $$\left(\frac{\sin x + 1}{\cos x}\right)^2 = 6$$. 4. **Simplify:** $$\frac{(\sin x + 1)^2}{\cos^2 x} = 6$$ Multiply both sides by $$\cos^2 x$$: $$(\sin x + 1)^2 = 6 \cos^2 x$$. 5. **Use the Pythagorean identity:** $$\cos^2 x = 1 - \sin^2 x$$, so: $$(\sin x + 1)^2 = 6(1 - \sin^2 x)$$ 6. **Expand and simplify:** $$(\sin x)^2 + 2 \sin x + 1 = 6 - 6 \sin^2 x$$ Bring all terms to one side: $$(\sin x)^2 + 2 \sin x + 1 - 6 + 6 \sin^2 x = 0$$ Combine like terms: $$7 \sin^2 x + 2 \sin x - 5 = 0$$ 7. **Let $$y = \sin x$$, solve quadratic:** $$7 y^2 + 2 y - 5 = 0$$ Use quadratic formula: $$y = \frac{-2 \pm \sqrt{2^2 - 4 \times 7 \times (-5)}}{2 \times 7} = \frac{-2 \pm \sqrt{4 + 140}}{14} = \frac{-2 \pm \sqrt{144}}{14} = \frac{-2 \pm 12}{14}$$ 8. **Calculate roots:** $$y_1 = \frac{-2 + 12}{14} = \frac{10}{14} = \frac{5}{7} \approx 0.714$$ $$y_2 = \frac{-2 - 12}{14} = \frac{-14}{14} = -1$$ 9. **Find $$x$$ values:** Since $$y = \sin x$$, - For $$y_1 = \frac{5}{7}$$, $$x = \arcsin\left(\frac{5}{7}\right)$$. - For $$y_2 = -1$$, $$x = \arcsin(-1) = -90^\circ$$ (not in $$0^\circ \leq x \leq 540^\circ$$, but we will consider the general solution). 10. **Recall $$x = 3\theta$$ and $$0^\circ \leq \theta \leq 180^\circ$$, so $$0^\circ \leq x \leq 540^\circ$$. 11. **Find all $$x$$ in $$[0^\circ, 540^\circ]$$ for each root:** - For $$\sin x = \frac{5}{7}$$, solutions are: $$x = \arcsin\left(\frac{5}{7}\right) \approx 45.58^\circ$$ and $$x = 180^\circ - 45.58^\circ = 134.42^\circ$$ Also add $$360^\circ$$ to these to get solutions in $$[0, 540]$$: $$x = 45.58^\circ + 360^\circ = 405.58^\circ$$ $$x = 134.42^\circ + 360^\circ = 494.42^\circ$$ - For $$\sin x = -1$$, solution in $$[0, 540]$$ is: $$x = 270^\circ$$ 12. **Convert back to $$\theta$$:** $$\theta = \frac{x}{3}$$ So the solutions are: $$\theta_1 = \frac{45.58^\circ}{3} \approx 15.19^\circ$$ $$\theta_2 = \frac{134.42^\circ}{3} \approx 44.81^\circ$$ $$\theta_3 = \frac{405.58^\circ}{3} \approx 135.19^\circ$$ $$\theta_4 = \frac{494.42^\circ}{3} \approx 164.81^\circ$$ $$\theta_5 = \frac{270^\circ}{3} = 90^\circ$$ 13. **Check all solutions are within $$0^\circ \leq \theta \leq 180^\circ$$, which they are.** **Final answer:** $$\boxed{\theta \approx 15.19^\circ, 44.81^\circ, 90^\circ, 135.19^\circ, 164.81^\circ}$$