Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>Which of the following set of values of <span>\(x\)</span> satisfies the equation <span>\(2^{\frac{\sin^2 x - 3\sin x + 1}{2\sin^2 x - 3\sin x + 1}} + 2^{\frac{2 - 2\sin^2 x + 3\sin x}{2\sin^2 x - 3\sin x + 1}} = 9\)</span></p>
<p>(a) <span>\(x = n\pi ± \frac{\pi}{6}\)</span>, <span>\(n \in \mathbb{I}\)</span></p>
<p>(b) <span>\(x = n\pi ± \frac{\pi}{3}\)</span>, <span>\(n \in \mathbb{I}\)</span></p>
<p>(c) <span>\(x = n\pi\)</span>, <span>\(n \in \mathbb{I}\)</span></p>
<p>(d) <span>\(x = 2n\pi + \frac{\pi}{2}\)</span>, <span>\(n \in \mathbb{I}\)</span></p>
Step-by-Step Solution
Key Concept: Use substitution to convert the equation into a quadratic form, then solve for the values of sine.
<p><strong>Step 1:</strong> Let <span>\(t = 2^{\frac{\sin^2 x - 3\sin x + 1}{2\sin^2 x - 3\sin x + 1}}\)</span>. Then <span>\(t + \frac{8}{t} = 9 \Rightarrow t^2 - 9t + 8 = 0 \Rightarrow t = 1, 8\)</span></p><p><strong>Step 2:</strong> If <span>\(t = 1\)</span>: <span>\(\frac{\sin^2 x - 3\sin x + 1}{2\sin^2 x - 3\sin x + 1} = 0 \Rightarrow \sin^2 x - 3\sin x + 1 = 0\)</span></p><p><strong>Step 3:</strong> If <span>\(t = 8\)</span>: <span>\(\frac{\sin^2 x - 3\sin x + 1}{2\sin^2 x - 3\sin x + 1} = 3 \Rightarrow 2\sin^2 x - 3\sin x + 1 = 0\)</span> or <span>\(2\sin^2 x - 3\sin x + 1 = 3\)</span></p><p><strong>Step 4:</strong> Solving: <span>\(\sin x = \frac{1}{2}, 1\)</span> gives <span>\(x = n\pi ± \frac{\pi}{6}\)</span> or <span>\(x = 2n\pi + \frac{\pi}{2}\)</span></p><p>∴ Answer is A,D.</p>
Correct Answer: A,D