AP Precalculus -3.10 Trigonometric Equations and Inequalities- MCQ Exam Style Questions - Effective Fall 2023
AP Precalculus -3.10 Trigonometric Equations and Inequalities- MCQ Exam Style Questions – Effective Fall 2023
AP Precalculus -3.10 Trigonometric Equations and Inequalities- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
▶️ Answer/Explanation
\( f(t) = \sin^2 t – 1 = 0 \)
→ \( \sin^2 t = 1 \)
→ \( \sin t = \pm 1 \)
This occurs when \( t = \frac{\pi}{2} + k\pi \) for integer \( k \).
Since \( k \) can be any integer, there are infinitely many solutions.
✅ Answer: (D)
▶️ Answer/Explanation
The equation \( \tan x = 3 \) has one principal solution \( x = \arctan 3 \).
Since the tangent function has period \( \pi \), the general solution is \( x = \arctan 3 + \pi k \), where \( k \) is any integer.
✅ Answer: (D)
▶️ Answer/Explanation
The table shows \( y \) values in \([-\frac{\pi}{2}, \frac{\pi}{2}]\) for \( x \) in \([-1, 1]\), and \( y = 0 \) when \( x = 0 \).
This matches the inverse sine function: \( y = \sin^{-1} x \) has domain \([-1, 1]\) and range \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
✅ Answer: (D)
▶️ Answer/Explanation
\( M(t) \) is a sine function with period \( 365 \) days, amplitude \( 150 \), midline \( 174 \).
Sine increases from minimum to maximum in the first half of its cycle, then decreases from maximum to minimum in the second half.
The maximum occurs when \( \frac{2\pi}{365}t = \frac{\pi}{2} \Rightarrow t = 91.25 \approx 91 \).
The minimum occurs when \( \frac{2\pi}{365}t = \frac{3\pi}{2} \Rightarrow t = 273.75 \approx 274 \).
Thus \( M \) is decreasing between these: approximately \( 92 < t < 273 \).
✅ Answer: (B)
▶️ Answer/Explanation
Rewrite: \(2\sin^2\theta + \sin\theta = 0\).
Factor: \(\sin\theta (2\sin\theta + 1) = 0\).
Case 1: \(\sin\theta = 0 \) ⇒ \(\theta = 0, \pi\) in \([0, 2\pi)\).
Case 2: \(2\sin\theta + 1 = 0 \) ⇒ \(\sin\theta = -\frac{1}{2}\) ⇒ \(\theta = \frac{7\pi}{6}, \frac{11\pi}{6}\).
All solutions: \(0, \pi, \frac{7\pi}{6}, \frac{11\pi}{6}\).
✅ Answer: (B)
▶️ Answer/Explanation
Given \( f(a) = b \) and \( f \) resembles \( y = \cos x \).
Then \( a = \cos^{-1} b \), i.e., \( (b, a) \) satisfies \( y = \cos^{-1} x \).
To have \((b, a)\) as a solution to a system, the second equation must force \( x = b \) and \( y = a \), which is exactly system (B).
✅ Answer: (B)
▶️ Answer/Explanation
Solve \( 5 + 7\cos x = 0 \) ⇒ \( \cos x = -\frac{5}{7} \).
The reference angle is \( \alpha = \arccos\left(\frac{5}{7}\right) \).
Since \( \cos x \) is negative in Quadrants II and III, and \( \pi < x < 2\pi \) is Quadrants III and IV, we need Quadrant III: \( x = \pi + \alpha \).
Thus \( x = \pi + \arccos\left(\frac{5}{7}\right) \).
✅ Answer: (B)
▶️ Answer/Explanation
Solve \( 1 + 3\sec x = -5 \) ⇒ \( 3\sec x = -6 \) ⇒ \( \sec x = -2 \).
Thus \( \cos x = -\frac{1}{2} \).
Solutions in \( [0, 2\pi) \): \( x = \frac{2\pi}{3}, \frac{4\pi}{3} \).
✅ Answer: (C)
▶️ Answer/Explanation
Expand: \( \cos(x + \frac{\pi}{6}) = \cos x\cos\frac{\pi}{6} – \sin x\sin\frac{\pi}{6} = \frac{\sqrt{3}}{2}\cos x – \frac{1}{2}\sin x \).
Set equal to \( \frac{1}{2} \):
\( \frac{\sqrt{3}}{2}\cos x – \frac{1}{2}\sin x = \frac{1}{2} \).
Multiply by 2: \( \sqrt{3}\cos x – \sin x = 1 \).
✅ Answer: (A)
▶️ Answer/Explanation
Use identity \( \sec^2 x = 1 + \tan^2 x \):
\( g(x) = (1 + \tan^2 x) + \tan x = 1 \) ⇒ \( \tan^2 x + \tan x = 0 \).
Factor: \( \tan x (\tan x + 1) = 0 \).
Case 1: \( \tan x = 0 \) ⇒ \( x = 0, \pi, 2\pi \).
Case 2: \( \tan x = -1 \) ⇒ \( x = \frac{3\pi}{4}, \frac{7\pi}{4} \).
All solutions in [0, 2π]: \( 0, \frac{3\pi}{4}, \pi, \frac{7\pi}{4}, 2\pi \).
✅ Answer: (A)
▶️ Answer/Explanation
Substitute \( \sin^2 x = 1 – \cos^2 x \):
\( f(x) = (1 – \cos^2 x) + \cos x + 1 = -\cos^2 x + \cos x + 2 \).
Set \( f(x) = 0 \): \( -\cos^2 x + \cos x + 2 = 0 \) ⇒ multiply by -1:
\( \cos^2 x – \cos x – 2 = 0 \).
✅ Answer: (B)
▶️ Answer/Explanation
\( 2\cos\theta > -1 \) ⇒ \( \cos\theta > -\frac{1}{2} \).
On \( [-\pi, \pi] \), this holds for \( -\frac{2\pi}{3} < \theta < \frac{2\pi}{3} \) (excluding endpoints where equality).
\( 2\sin\theta > \sqrt{3} \) ⇒ \( \sin\theta > \frac{\sqrt{3}}{2} \).
On \( [-\pi, \pi] \), this holds for \( \frac{\pi}{3} < \theta < \frac{2\pi}{3} \).
Intersection of both inequalities: \( \frac{\pi}{3} < \theta < \frac{2\pi}{3} \).
✅ Answer: (D)
▶️ Answer/Explanation
\( \tan\theta = 1 \) ⇒ \( \theta = \frac{\pi}{4} + \pi k \).
On \( [0, 2\pi) \): \( \frac{\pi}{4} \) (QI) and \( \frac{5\pi}{4} \) (QIII).
✅ Answer: (B)
▶️ Answer/Explanation
1. Solve for \(\sin\theta\):
\(2\sin\theta = -1 \implies \sin\theta = -\frac{1}{2}\)
2. Identify Quadrants:
Sine is negative in Quadrants III and IV.
3. Find Angles:
Reference angle is \(\frac{\pi}{6}\) (since \(\sin\frac{\pi}{6}=\frac{1}{2}\)).
Quadrant III: \(\pi + \frac{\pi}{6} = \frac{7\pi}{6}\)
Quadrant IV: \(2\pi – \frac{\pi}{6} = \frac{11\pi}{6}\)
✅ Answer: (D)
▶️ Answer/Explanation
Rewrite:
\[ 2\sin^2\theta + \sin\theta = 0 \]
\[ \sin\theta(2\sin\theta + 1) = 0 \]
Case 1: \( \sin\theta = 0 \) ⇒ \( \theta = 0, \pi \) in \( [0, 2\pi) \).
Case 2: \( 2\sin\theta + 1 = 0 \) ⇒ \( \sin\theta = -\frac{1}{2} \) ⇒ \( \theta = \frac{7\pi}{6}, \frac{11\pi}{6} \).
All solutions: \( 0, \pi, \frac{7\pi}{6}, \frac{11\pi}{6} \).
✅ Answer: (B)
▶️ Answer/Explanation
\( g(x) = \sin(2.25x + 0.2) + 0.5 \).
Set \( g(x) = 0 \):
\[ \sin(2.25x + 0.2) = -0.5 \]
General solution: \( 2.25x + 0.2 = \frac{7\pi}{6} + 2n\pi \) or \( \frac{11\pi}{6} + 2n\pi \).
Solve for \( x \) in \( [0,\pi] \):
Using calculator approximations:
For \( n = 0 \):
• \( 2.25x + 0.2 = 7\pi/6 \approx 3.66519 \) ⇒ \( x \approx 1.540 \)
• \( 2.25x + 0.2 = 11\pi/6 \approx 5.75959 \) ⇒ \( x \approx 2.471 \)
Both are in \( [0,\pi] \).
✅ Answer: (C)