Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>If <i>A</i> and <i>B</i> are acute positive angles satisfying the equations \(3\sin 2A + 2\sin^2 B = 1\) and \(3\sin 2A - 2\sin 3B = 0\), then \(A + 2B\) is equal to</p>
<p>(a) \(\frac{\pi}{2}\)</p>
<p>(b) \(\frac{5\pi}{2}\)</p>
<p>(c) \(\frac{\pi}{3}\)</p>
<p>(d) \(\frac{2\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: Solve the system of trigonometric equations simultaneously to find the relationship between angles A and B, then compute their sum.
<p>From the given equations: $3\sin 2A + 2\sin^2 B = 1$ and $3\sin 2A - 2\sin 3B = 0$</p><p>Solving these simultaneous equations by standard trigonometric methods yields $A + 2B = \frac{\pi}{2}$.</p>
Correct Answer: A