1. **State the problem:** Solve the trigonometric equation $$7 \cos 2x - 24 \sin 2x = 12.5$$ for $$0 \leq x < 180^\circ$$, rounding answers to 1 decimal place. 2. **Use the formula for combining sine and cosine:** Any expression of the form $$a \cos \theta + b \sin \theta$$ can be rewritten as $$R \cos(\theta - \phi)$$ where $$R = \sqrt{a^2 + b^2}$$ and $$\tan \phi = \frac{b}{a}$$. 3. **Calculate $$R$$ and $$\phi$$:** $$R = \sqrt{7^2 + (-24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25$$ $$\tan \phi = \frac{-24}{7}$$ so $$\phi = \arctan\left(\frac{-24}{7}\right) \approx -73.74^\circ$$ 4. **Rewrite the equation:** $$7 \cos 2x - 24 \sin 2x = 25 \cos(2x - (-73.74^\circ)) = 25 \cos(2x + 73.74^\circ)$$ 5. **Set equal to 12.5:** $$25 \cos(2x + 73.74^\circ) = 12.5$$ Divide both sides by 25: $$\cancel{25} \cos(2x + 73.74^\circ) = \frac{12.5}{\cancel{25}}$$ $$\cos(2x + 73.74^\circ) = 0.5$$ 6. **Solve for the angle:** $$2x + 73.74^\circ = \cos^{-1}(0.5)$$ The principal values of $$\cos^{-1}(0.5)$$ are $$60^\circ$$ and $$300^\circ$$. 7. **Find $$x$$ values:** For $$2x + 73.74^\circ = 60^\circ$$: $$2x = 60^\circ - 73.74^\circ = -13.74^\circ$$ $$x = -6.87^\circ$$ (not in $$0 \leq x < 180^\circ$$, discard) For $$2x + 73.74^\circ = 300^\circ$$: $$2x = 300^\circ - 73.74^\circ = 226.26^\circ$$ $$x = 113.13^\circ$$ (valid) 8. **Use periodicity:** Since cosine has period $$360^\circ$$, for $$2x + 73.74^\circ = 60^\circ + 360^\circ k$$ and $$300^\circ + 360^\circ k$$, where $$k$$ is an integer. Check for $$k=1$$: $$2x + 73.74^\circ = 60^\circ + 360^\circ = 420^\circ$$ $$2x = 420^\circ - 73.74^\circ = 346.26^\circ$$ $$x = 173.13^\circ$$ (valid) For $$k=1$$ in second solution: $$2x + 73.74^\circ = 300^\circ + 360^\circ = 660^\circ$$ $$2x = 660^\circ - 73.74^\circ = 586.26^\circ$$ $$x = 293.13^\circ$$ (not in range) 9. **Final solutions:** $$x \approx 113.1^\circ, 173.1^\circ$$ --- **Second problem:** Find a complete equation for the alternative model given $$H = 60 - 0.145(t - 20)^2$$ and $$H = 29 \cos(9t + \alpha) + \beta$$ 1. The quadratic model has vertex at $$t=20$$ with height $$H=60$$. 2. The cosine model has amplitude $$29$$, so the vertical shift $$\beta$$ is the average height, which equals the vertex height: $$\beta = 60$$ 3. The period of the cosine is $$\frac{360^\circ}{9} = 40$$ seconds. 4. The vertex corresponds to the maximum height, so the phase shift $$\alpha$$ satisfies: $$9 \times 20 + \alpha = 0^\circ$$ (cosine max at 0°) Solve for $$\alpha$$: $$\alpha = -180^\circ$$ Since $$0 \leq \alpha < 360^\circ$$, add 360°: $$\alpha = 180^\circ$$ 5. **Complete alternative model:** $$H = 29 \cos(9t + 180^\circ) + 60$$ --- **Summary:** - Solutions for $$x$$: $$113.1^\circ$$ and $$173.1^\circ$$ - Alternative model: $$H = 29 \cos(9t + 180^\circ) + 60$$