1. **State the problem:** We need to show that the equation $$\tan 2x = 5 \sin 2x$$ can be written as $$(1 - 5 \cos 2x) \sin 2x = 0$$ and then solve this equation for $0 \leq x \leq 180^\circ$. 2. **Rewrite the tangent function:** Recall that $$\tan 2x = \frac{\sin 2x}{\cos 2x}$$. 3. **Substitute into the equation:** $$\frac{\sin 2x}{\cos 2x} = 5 \sin 2x$$ 4. **Multiply both sides by $$\cos 2x$$ to clear the denominator:** $$\sin 2x = 5 \sin 2x \cos 2x$$ 5. **Bring all terms to one side:** $$\sin 2x - 5 \sin 2x \cos 2x = 0$$ 6. **Factor out $$\sin 2x$$:** $$(1 - 5 \cos 2x) \sin 2x = 0$$ This completes part (a). --- 7. **Solve the equation $$(1 - 5 \cos 2x) \sin 2x = 0$$ for $0 \leq x \leq 180^\circ$.** 8. **Set each factor equal to zero:** - Case 1: $$\sin 2x = 0$$ - Case 2: $$1 - 5 \cos 2x = 0$$ or $$\cos 2x = \frac{1}{5}$$ 9. **Solve Case 1:** $$\sin 2x = 0$$ implies $$2x = n\times 180^\circ$$ for integer $$n$$. Within $$0 \leq x \leq 180^\circ$$, corresponding to $$0 \leq 2x \leq 360^\circ$$, we have $$2x = 0^\circ, 180^\circ, 360^\circ$$ giving: $$x = 0^\circ, 90^\circ, 180^\circ$$ 10. **Solve Case 2:** $$\cos 2x = \frac{1}{5} = 0.2$$ Use the inverse cosine: $$2x = \cos^{-1}(0.2) = 78.463^\circ$$ (approx) Also, $$\cos \theta = \cos (360^\circ - \theta)$$, so $$2x = 360^\circ - 78.463^\circ = 281.537^\circ$$ Within $$0 \leq 2x \leq 360^\circ$$ these are the two solutions. Divide by 2: $$x = 39.2^\circ, 140.8^\circ$$ (rounded to 1 decimal place) 11. **Final solutions:** $$x = 0^\circ, 39.2^\circ, 90^\circ, 140.8^\circ, 180^\circ$$ These are all the solutions to the original equation for $0 \leq x \leq 180^\circ$. **Answers are given to 1 decimal place as requested.**