利用 $R \cos(\theta - \alpha)$（其中 $R > 0$ 且 $0^\circ \le \alpha \le 90^\circ$），求解方程
$5 \cos \theta + 8 \sin \theta = 6$
其中 $0^\circ \le \theta \le 180^\circ$。 写出 $20 - (5 \cos x + 8 \sin x)^2$ 的最大值，并求出在 $0^\circ \le x \le 180^\circ$ 范围内对应的 $x$ 值。

1. Express $5 \cos \theta + 8 \sin \theta$ in the form $R \cos(\theta - \alpha)$.

Let $5 \cos \theta + 8 \sin \theta = R \cos(\theta - \alpha) = R \cos \theta \cos \alpha + R \sin \theta \sin \alpha$.
Comparing coefficients:
$R \cos \alpha = 5$
$R \sin \alpha = 8$
$R = \sqrt{5^2 + 8^2} = \sqrt{89}$
$\tan \alpha = \frac{8}{5} \implies \alpha = \tan^{-1}(1.6) \approx 57.9946^\circ$
So, $5 \cos \theta + 8 \sin \theta = \sqrt{89} \cos(\theta - 57.9946^\circ)$.

2. Set up the equation and find the basic angle.

$\sqrt{89} \cos(\theta - 57.9946^\circ) = 6$
$\cos(\theta - 57.9946^\circ) = \frac{6}{\sqrt{89}}$
Basic angle $= \cos^{-1}\left(\frac{6}{\sqrt{89}}\right) \approx 50.506^\circ$

3. Solve for $\theta$ in the given interval $0^\circ \le \theta \le 180^\circ$.

The interval for $\theta - 57.9946^\circ$ is $-57.9946^\circ \le \theta - 57.9946^\circ \le 122.005^\circ$.
$\theta - 57.9946^\circ = -50.506^\circ \text{ or } 50.506^\circ$
$\theta = 57.9946^\circ - 50.506^\circ = 7.4886^\circ \approx 7.5^\circ$
$\theta = 57.9946^\circ + 50.506^\circ = 108.5006^\circ \approx 108.5^\circ$

1. Express the given expression using the R-formula result from part (i)

From part (i), $5 \cos x + 8 \sin x = \sqrt{89} \cos(x - 57.99^\circ)$.
Substitute this into the expression:
$20 - (5 \cos x + 8 \sin x)^2 = 20 - (\sqrt{89} \cos(x - 57.99^\circ))^2$
$= 20 - 89 \cos^2(x - 57.99^\circ)$

2. Find the maximum value of the expression

To maximize $20 - 89 \cos^2(x - 57.99^\circ)$, we must minimize $89 \cos^2(x - 57.99^\circ)$.
Since $\cos^2(x - 57.99^\circ) \ge 0$, the minimum value is $0$.
Therefore, the largest value of the expression is $20 - 0 = 20$.

3. Find the corresponding value of $x$

The maximum occurs when $\cos(x - 57.9946^\circ) = 0$.
$x - 57.9946^\circ = 90^\circ$ (since $0^\circ \le x \le 180^\circ$)
$x = 90^\circ + 57.9946^\circ = 147.9946^\circ \approx 148.0^\circ$ (to 1 d.p.)