Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>If <span class="math">\sin\theta + \sqrt{3}\cos\theta = 6x - x^2 - 11</span>, <span class="math">0 \leq \theta \leq 4\pi</span>, <span class="math">x \in \mathbb{R}</span>, then:</p>
<p>(a) no values of <span class="math">x</span> and <span class="math">\theta</span></p>
<p>(b) one value of <span class="math">x</span> and two values of <span class="math">\theta</span></p>
<p>(c) two values of <span class="math">x</span> and two values of <span class="math">\theta</span></p>
<p>(d) two pairs of values of <span class="math">(x, \theta)</span></p>
Step-by-Step Solution
Key Concept: Express the trigonometric expression in amplitude-phase form and the quadratic as a perfect square to find where they can simultaneously achieve their extreme values.
<p><strong>Solution:</strong> The left side <span class="math">\sin\theta + \sqrt{3}\cos\theta</span> can be written as <span class="math">2\sin(\theta + \frac{\pi}{3})</span>, which has maximum value 2 and minimum value -2. The right side <span class="math">6x - x^2 - 11 = -(x^2 - 6x + 11) = -(x-3)^2 - 2</span> has maximum value -2 (when <span class="math">x = 3</span>). For the equation to hold, both sides must equal -2. This gives <span class="math">x = 3</span> and <span class="math">\sin(\theta + \frac{\pi}{3}) = -1</span>, which yields specific values of <span class="math">\theta</span> in <span class="math">[0, 4\pi]</span>.</p><p>∴ Answer is (b) or (d).</p>
Correct Answer: B, D