AP Precalculus: Trigonometric Equations (Drill 1)

Question 1. Which of the following gives all solutions to \( 2\sin^2\theta = -\sin\theta \) on \( [0, 2\pi) \)?

• A) \( \theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6} \)
• B) \( \theta = \pi, \dfrac{7\pi}{6}, \dfrac{11\pi}{6} \)
• C) \( \theta = 0, \pi, \dfrac{7\pi}{6}, \dfrac{11\pi}{6} \) ✓
• D) \( \theta = 0, \dfrac{\pi}{6}, \pi, \dfrac{7\pi}{6} \) on the interval stated in the question

Explanation: Choice C is correct. Rewrite as \( 2\sin^2\theta + \sin\theta = 0 \) and factor: \( \sin\theta(2\sin\theta + 1) = 0 \). Setting \( \sin\theta = 0 \) gives \( \theta = 0 \) and \( \theta = \pi \). Setting \( \sin\theta = -\dfrac{1}{2} \) gives \( \theta = \dfrac{7\pi}{6} \) and \( \theta = \dfrac{11\pi}{6} \). All four values lie in \( [0, 2\pi) \). Choice A is incorrect because it omits both \( \theta = 0 \) and \( \theta = \pi \) from the factor \( \sin\theta = 0 \). Choice B is incorrect because it includes \( \theta = \pi \) but omits \( \theta = 0 \); both arise from \( \sin\theta = 0 \) and both are valid. Choice D is incorrect because \( \dfrac{\pi}{6} \) gives \( \sin(\pi/6) = \dfrac{1}{2} \), not \( -\dfrac{1}{2} \); the student used the correct reference angle but the wrong sign.

Question 2. If \( \sec\theta = -2 \) and \( \dfrac{\pi}{2} < \theta < \pi \), what is the value of \( \sin\theta \)?

• A) \( -\dfrac{\sqrt{3}}{2} \)
• B) \( -\dfrac{1}{2} \)
• C) \( \dfrac{1}{2} \)
• D) \( \dfrac{\sqrt{3}}{2} \) ✓

Explanation: Choice D is correct. Since \( \sec\theta = -2 \), we have \( \cos\theta = -\dfrac{1}{2} \). The reference angle satisfying \( |\cos\theta| = \dfrac{1}{2} \) is \( \dfrac{\pi}{3} \). In Quadrant II, \( \theta = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3} \), and \( \sin\!\left(\dfrac{2\pi}{3}\right) = \dfrac{\sqrt{3}}{2} \). Choice A is incorrect because \( -\dfrac{\sqrt{3}}{2} \) is the sine value in Quadrant III, where both sine and cosine are negative; the given restriction places \( \theta \) in Quadrant II, where sine is positive. Choice B is incorrect because \( -\dfrac{1}{2} \) is the value of \( \cos\theta \), not \( \sin\theta \). Choice C is incorrect because \( \dfrac{1}{2} \) would correspond to a reference angle of \( \dfrac{\pi}{6} \), but the reference angle here is \( \dfrac{\pi}{3} \).

Question 3. The table below shows selected values of \( g(x) = \sin(2.25x + 0.2) + 0.5 \) on the interval \( [0, \pi] \).

x	0	1	1.5	2	2.5	3
g(x)	1.198	0.270	−0.308	−0.387	0.297	0.997

Based on the table, on which of the following intervals does \( g \) have a zero?

• A) \( (0, 1) \) only
• B) \( (1, 1.5) \) and \( (2, 2.5) \) ✓
• C) \( (1, 1.5) \) only
• D) \( (1.5, 2) \) and \( (2.5, 3) \)

Explanation: Choice B is correct. A zero occurs wherever \( g \) changes sign. Between \( x = 1 \) and \( x = 1.5 \), \( g \) changes from \( 0.270 \) to \( -0.308 \), confirming a zero in \( (1, 1.5) \). Between \( x = 2 \) and \( x = 2.5 \), \( g \) changes from \( -0.387 \) to \( 0.297 \), confirming a zero in \( (2, 2.5) \). Choice A is incorrect because \( g(0) = 1.198 \) and \( g(1) = 0.270 \) are both positive; the function does not change sign on \( (0, 1) \). Choice C is incorrect because it identifies only the first sign change and misses the second. Choice D is incorrect because the function does not change sign within those intervals: \( g(1.5) \) and \( g(2) \) are both negative, and \( g(2.5) \) and \( g(3) \) are both positive.

Question 4. A Ferris wheel's height above the ground (in feet) is modeled by \( h(t) = 40\sin\!\left(\dfrac{\pi}{10}t\right) + 45 \), where \( t \) is time in seconds. During the first full revolution (\( 0 \leq t \leq 20 \)), for how many seconds is the rider at a height greater than 65 feet?

• A) \( \dfrac{10}{3} \)
• B) \( \dfrac{20}{3} \) ✓
• C) \( \dfrac{10\pi}{3} \)
• D) 10

Explanation: Choice B is correct. Set \( 40\sin\!\left(\dfrac{\pi}{10}t\right) + 45 > 65 \), giving \( \sin\!\left(\dfrac{\pi}{10}t\right) > \dfrac{1}{2} \). Let \( u = \dfrac{\pi}{10}t \); on one full period \( [0, 2\pi] \), \( \sin u > \dfrac{1}{2} \) when \( u \in \left(\dfrac{\pi}{6}, \dfrac{5\pi}{6}\right) \). Converting back: \( \dfrac{\pi}{6} < \dfrac{\pi}{10}t < \dfrac{5\pi}{6} \), so \( \dfrac{5}{3} < t < \dfrac{25}{3} \). Duration: \( \dfrac{25}{3} - \dfrac{5}{3} = \dfrac{20}{3} \) seconds. Choice A is incorrect because \( \dfrac{10}{3} \) is only half the interval width; the student divided by 2 rather than computing the full duration. Choice C is incorrect because the student left the expression in terms of \( \pi \) without completing the substitution; the factor of \( \pi \) cancels when converting from \( u \) back to \( t \). Choice D is incorrect because the rider spends half the revolution above the midline of 45 ft, not above 65 ft; the threshold of 65 ft corresponds to a narrower interval than half the period.

Question 5. How many solutions does \( 2\cos^2\theta - \cos\theta - 1 = 0 \) have on the interval \( [0, 2\pi) \)?

• A) 3 ✓
• B) 2
• C) 4
• D) 1

Explanation: Choice A is correct. Factor the quadratic in \( \cos\theta \): \( (2\cos\theta + 1)(\cos\theta - 1) = 0 \). From \( \cos\theta = 1 \): \( \theta = 0 \). From \( \cos\theta = -\dfrac{1}{2} \): \( \theta = \dfrac{2\pi}{3} \) and \( \theta = \dfrac{4\pi}{3} \). That gives 3 solutions in \( [0, 2\pi) \). Choice B is incorrect because the student found only the two solutions from \( \cos\theta = -\dfrac{1}{2} \) and overlooked \( \theta = 0 \) from the factor \( \cos\theta - 1 = 0 \). Choice C is incorrect because the student treated \( \cos\theta = 1 \) as yielding two solutions (\( \theta = 0 \) and \( \theta = 2\pi \)), but \( 2\pi \) is not in the half-open interval \( [0, 2\pi) \). Choice D is incorrect because the student found only \( \theta = 0 \) and stopped without fully factoring the quadratic.