Question - Trigonometric Equation Solving
Solution:
\[
\begin{align*}
&\text{给定方程式为} \quad 2\tan(\theta) = 3\cos(\theta) \\
&\text{解这个方程式，我们首先将其转换成同一个三角函数，} \\
&\text{使用基本恒等式} \quad \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}: \\
&2\frac{\sin(\theta)}{\cos(\theta)} = 3\cos(\theta) \\
&2\sin(\theta) = 3\cos^2(\theta) \\
&\text{因为} \quad \cos^2(\theta) = 1 - \sin^2(\theta), \text{代入上面方程得到} \\
&2\sin(\theta) = 3(1 - \sin^2(\theta)) \\
&3\sin^2(\theta) + 2\sin(\theta) - 3 = 0 \\
&\text{将}\sin(\theta)\text{代换为} x \\
&3x^2 + 2x - 3 = 0 \\
&\text{使用求根公式得到:} \\
&x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
&x = \frac{-2 \pm \sqrt{(2)^2 - 4(3)(-3)}}{2(3)} \\
&x = \frac{-2 \pm \sqrt{4 + 36}}{6} \\
&x = \frac{-2 \pm \sqrt{40}}{6} \\
&x = \frac{-2 \pm 2\sqrt{10}}{6} \\
&\text{因为} -1 \leq \sin(\theta) \leq 1, \text{所以我们舍弃} x > 1 \text{和} x < -1 \text{的根。} \\
&x = -1 \text{（不合理）或 } x = \frac{\sqrt{10}-1}{3} \\
&\text{所以}\sin(\theta) = \frac{\sqrt{10}-1}{3} \\
&\text{利用计算器得到} \theta \text{的两个可能的值(确保在 0° 至 360°之间)} \\
&\theta \approx 50.2^\circ \text{或} 360^\circ - 50.2^\circ \approx 309.8^\circ \\
&\text{最终答案为：} \theta \approx 50.2^\circ \text{或} 309.8^\circ
\end{align*}
\]
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