求介于 $0^\circ$ 和 $180^\circ$ 之间满足 $\tan^2\theta = 5 - \sec\theta$ 的 $\theta$ 的值。 (b) 求介于 $0^\circ$ 和 $360^\circ$ 之间满足方程 $6\cos\theta + 7\sin\theta = 4$ 的 $\theta$ 的值。

1. Use the trigonometric identity relating $\tan^2\theta$ and $\sec^2\theta$ to form a quadratic equation in $\sec\theta$.

Using $\tan^2\theta = \sec^2\theta - 1$,
$\sec^2\theta - 1 = 5 - \sec\theta$
$\sec^2\theta + \sec\theta - 6 = 0$

2. Factorise and solve the quadratic equation for $\sec\theta$.

$(\sec\theta + 3)(\sec\theta - 2) = 0$
$\sec\theta = -3 \quad \text{or} \quad \sec\theta = 2$

3. Find the values of $\theta$ in the range $0^\circ < \theta < 180^\circ$.

$\cos\theta = -\frac{1}{3} \quad \text{or} \quad \cos\theta = \frac{1}{2}$
For $\cos\theta = \frac{1}{2}$, $\theta = 60^\circ$.
For $\cos\theta = -\frac{1}{3}$, $\theta = 180^\circ - 70.53^\circ = 109.47^\circ \approx 109.5^\circ$.

4. Express $6\cos\theta + 7\sin\theta$ in the form $R\cos(\theta - \alpha)$.

Let $6\cos\theta + 7\sin\theta = R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$.
Comparing coefficients:
$R\cos\alpha = 6$
$R\sin\alpha = 7$
$R = \sqrt{6^2 + 7^2} = \sqrt{85} \approx 9.2195$
$\tan\alpha = \frac{7}{6} \implies \alpha \approx 49.399^\circ$

5. Solve the equation for the basic angle.

$\sqrt{85}\cos(\theta - 49.399^\circ) = 4$
$\cos(\theta - 49.399^\circ) = \frac{4}{\sqrt{85}}$
$\text{Basic angle} = \cos^{-1}\left(\frac{4}{\sqrt{85}}\right) \approx 64.290^\circ$

6. Find the values of $\theta$ in the range $0^\circ \le \theta \le 360^\circ$.

$\theta - 49.399^\circ = 64.290^\circ \quad \text{or} \quad -64.290^\circ$
$\theta = 49.399^\circ + 64.290^\circ = 113.689^\circ \approx 113.7^\circ$
$\theta = 49.399^\circ - 64.290^\circ = -14.891^\circ$
Since $\theta$ must be between $0^\circ$ and $360^\circ$, we add $360^\circ$ to the negative value:
$\theta = -14.891^\circ + 360^\circ = 345.109^\circ \approx 345.1^\circ$