Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>The general solution of \(\sin 2\theta \sec\theta + \sqrt{3}\tan\theta = 0\) is</p>
<p>(a) \(\theta = n\pi + (-1)^{n+1} \frac{\pi}{3}, \theta = n\pi, n \in \mathbb{I}\)</p>
<p>(b) Other options</p>
<p>(c) Other options</p>
<p>(d) Other options</p>
Step-by-Step Solution
Key Concept: Simplify using $\sec\theta = \frac{1}{\cos\theta}$ and factor out common trigonometric terms.
<p>Factoring: $\sin 2\theta \sec\theta + \sqrt{3}\tan\theta = \frac{2\sin\theta\cos\theta}{\cos\theta} + \sqrt{3}\tan\theta = 2\sin\theta + \sqrt{3}\tan\theta = 0$. This gives $\tan\theta(\sqrt{3} + \frac{2\cos\theta}{\cos\theta}) = 0$, yielding $\theta = n\pi$ or $\theta = n\pi + (-1)^{n+1}\frac{\pi}{3}$.</p>
Correct Answer: a