Trigonometry & Inverse Trigonometry
General Solutions of Trigonometric Equations
Grade 11
Question:
<p>If \(2\sin\theta + 1 = 0\) and \(\sqrt{3}\tan\theta = 1\) then the most general value of \(\theta\) is</p>
<p>(a) \(n\pi \pm \dfrac{\pi}{6}\)</p>
<p>(b) \(n\pi + (-1)^n \dfrac{\pi}{6}\)</p>
<p>(c) \(2n\pi + \dfrac{7\pi}{6}\)</p>
<p>(d) \(2n\pi + \dfrac{11\pi}{6}\)</p>
Step-by-Step Solution
Key Concept: Find the specific angle(s) satisfying BOTH equations simultaneously, then express the general solution using the periodicity of that unique angle.
<p><strong>Step 1:</strong> From the first equation: $2\sin\theta + 1 = 0$ → $\sin\theta = -\frac{1}{2}$</p><p>This gives $\theta = -\frac{\pi}{6}$ or $\theta = -\frac{5\pi}{6}$ (in principal range)</p><p><strong>Step 2:</strong> From the second equation: $\sqrt{3}\tan\theta = 1$ → $\tan\theta = \frac{1}{\sqrt{3}}$</p><p>This gives $\theta = \frac{\pi}{6}$ or $\theta = \frac{\pi}{6} + \pi$ (in principal range)</p><p><strong>Step 3:</strong> Check which angle satisfies BOTH equations:</p><p>• At $\theta = -\frac{\pi}{6}$: $\sin(-\frac{\pi}{6}) = -\frac{1}{2}$ ✓ and $\tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$ ✗</p><p>• At $\theta = -\frac{5\pi}{6}$: $\sin(-\frac{5\pi}{6}) = -\frac{1}{2}$ ✓ and $\tan(-\frac{5\pi}{6}) = \frac{1}{\sqrt{3}}$ ✓</p><p><strong>Step 4:</strong> The unique solution is $\theta = -\frac{5\pi}{6}$ (or equivalently $\frac{7\pi}{6}$)</p><p><strong>Step 5:</strong> Apply periodicity of tangent (since $\tan\theta = \frac{1}{\sqrt{3}}$ and $\sin\theta = -\frac{1}{2}$ both have period related constraints): The general solution is $\theta = n\pi - \frac{5\pi}{6}$ where $n \in \mathbb{Z}$</p><p>∴ Answer: C</p>
Correct Answer: C