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.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The function $ f $ is given by $ f(t) = \sin^2 t – 1 $. For how many values of $ t $ does $ f(t) = 0 $?

(A) None
(B) One
(C) Two
(D) Infinitely many

▶️ Answer/Explanation

Detailed solution

$ 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)

Question

The function $ g $ is given by $ g(x) = \tan x $. What are all solutions to $ g(x) = 3 $?

(A) $ x = \arctan 3 $ and $ x = \pi + \arctan 3 $ only
(B) $ x = \arctan 3 $ and $ x = \pi + \mathrm{arctanh} (3 + \pi) $ only
(C) $ x = 2\pi k + \arctan 3 $ only, where $ k $ is any integer
(D) $ x = \pi k + \arctan 3 $, where $ k $ is any integer

▶️ Answer/Explanation

Detailed solution

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)

Question

The table gives ordered pairs $(x, y)$ that are solutions to which of the following?

$x$	$y$
$-1$	$-\frac{\pi}{2}$
$-\frac{1}{2}$	$-\frac{\pi}{6}$
0	0
$\frac{1}{2}$	$\frac{\pi}{6}$
1	$\frac{\pi}{2}$

(A) $ y = \cos x $
(B) $ y = \cos^{-1} x $
(C) $ y = \sin x $
(D) $ y = \sin^{-1} x $

▶️ Answer/Explanation

Detailed solution

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)

Question

The number of moose in a park is modeled by $ M(t) = 174 + 150 \sin\left(\frac{2\pi}{365}t\right) $, where $ t $ is days since January 1. Which statement is true about when $ M $ is decreasing?

(A) $ 0 < t < 182 $
(B) $ 92 < t < 273 $
(C) $ 183 < t < 365 $
(D) $ 0 < t < 91 $ and $ 274 < t < 365 $

▶️ Answer/Explanation

Detailed solution

$ 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)

Question

What are all values of $\theta$, for $0 \leq \theta < 2\pi$, where $2\sin^2\theta = -\sin\theta$?

(A) $0, \pi, \frac{\pi}{6}, \text{ and } \frac{5\pi}{6}$
(B) $0, \pi, \frac{7\pi}{6}, \text{ and } \frac{11\pi}{6}$
(C) $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{3}, \text{ and } \frac{5\pi}{3}$
(D) $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{2\pi}{3}, \text{ and } \frac{4\pi}{3}$

▶️ Answer/Explanation

Detailed solution

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)

Question

The graphs of a trigonometric function $ f $ and a line are given. The point of intersection $(a, b)$ occurs where $ f(x) = b $. For which of the following systems could $(b, a)$ be a solution?

(A) $\begin{cases} y = \cos^{-1}x \\ y = b \end{cases}$
(B) $\begin{cases} y = \cos^{-1}x \\ x = b \end{cases}$
(C) $\begin{cases} y = \sin^{-1}x \\ y = b \end{cases}$
(D) $\begin{cases} y = \sin^{-1}x \\ x = b \end{cases}$

▶️ Answer/Explanation

Detailed solution

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)

Question

The function $ f(x) = 5 + 7\cos x $. For what value of $ x $ on $ \pi < x < 2\pi $ does $ f(x) = 0 $?

(A) $ \arccos\left(-\frac{5}{7}\right) $
(B) $ \pi + \arccos\left(\frac{5}{7}\right) $
(C) $ \pi + \arccos\left(-\frac{5}{7}\right) $
(D) $ 2\pi – \arccos\left(\frac{5}{7}\right) $

▶️ Answer/Explanation

Detailed solution

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)

Question

Let $ f(x) = 1 + 3\sec x $ and $ g(x) = -5 $. What are the $x$-coordinates of the intersections for $ 0 \leq x < 2\pi $?

(A) $ \frac{\pi}{3}, \frac{5\pi}{3} $
(B) $ \frac{\pi}{6}, \frac{5\pi}{6} $
(C) $ \frac{2\pi}{3}, \frac{4\pi}{3} $
(D) $ \frac{7\pi}{6}, \frac{11\pi}{6} $

▶️ Answer/Explanation

Detailed solution

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)

Question

The function $ f(x) = \cos(x + \frac{\pi}{6}) $. The solutions to which equation on $ 0 \leq x \leq 2\pi $ are the solutions to $ f(x) = \frac{1}{2} $?

(A) $ \sqrt{3}\cos x – \sin x = 1 $
(B) $ \sqrt{3}\cos x + \sin x = 1 $
(C) $ \sqrt{3}\sin x – \cos x = 1 $
(D) $ \sqrt{3}\sin x + \cos x = 1 $

▶️ Answer/Explanation

Detailed solution

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)

Question

The function $ g(x) = \sec^2 x + \tan x $. What are all solutions to $ g(x) = 1 $ on $ 0 \leq x \leq 2\pi $?

(A) $ x = 0, \frac{3\pi}{4}, \pi, \frac{7\pi}{4}, 2\pi $
(B) $ x = \frac{\pi}{4}, \frac{\pi}{2}, \frac{5\pi}{4}, \frac{3\pi}{2} $
(C) $ x = \pi k $ and $ x = -\frac{\pi}{4} + \pi k $ for any integer $ k $
(D) $ x = \frac{\pi}{2} + \pi k $ and $ x = \frac{\pi}{4} + \pi k $ for any integer $ k $

▶️ Answer/Explanation

Detailed solution

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)

Question

The function $ f(x) = \sin^2 x + \cos x + 1 $. The solutions to which equation are also solutions to $ f(x) = 0 $?

(A) $ \cos^2 x + \cos x = 0 $
(B) $ \cos^2 x – \cos x – 2 = 0 $
(C) $ \cos^2 x + \cos x + 2 = 0 $
(D) $ \sin^2 x – \sin x + 2 = 0 $

▶️ Answer/Explanation

Detailed solution

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)

Question

What are all values of $ \theta $ in $ -\pi \leq \theta \leq \pi $ for which $ 2\cos\theta > -1 $ and $ 2\sin\theta > \sqrt{3} $?

(A) $ -\frac{5\pi}{6} < \theta < \frac{5\pi}{6} $
(B) $ \frac{\pi}{6} < \theta < \frac{5\pi}{6} $ only
(C) $ -\frac{2\pi}{3} < \theta < \frac{2\pi}{3} $ only
(D) $ \frac{\pi}{3} < \theta < \frac{2\pi}{3} $ only

▶️ Answer/Explanation

Detailed solution

$ 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)

Question

On $ 0 \leq \theta \leq 2\pi $, if $ \tan\theta = 1 $, which statement is true?

(A) Only Quadrant I
(B) Quadrants I and III only
(C) All four quadrants
(D) No value satisfies

▶️ Answer/Explanation

Detailed solution

$ \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)

Question

The function $k$ is given by $k(\theta)=2\sin\theta$. What are all values of $\theta$, for $0\le\theta<2\pi$, where $k(\theta)=-1$?

(A) $\theta=\frac{\pi}{6}$ and $\theta=\frac{5\pi}{6}$
(B) $\theta=\frac{\pi}{3}$ and $\theta=\frac{5\pi}{3}$
(C) $\theta=\frac{2\pi}{3}$ and $\theta=\frac{4\pi}{3}$
(D) $\theta=\frac{7\pi}{6}$ and $\theta=\frac{11\pi}{6}$

▶️ Answer/Explanation

Detailed solution

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)

Question

What are all values of $ \theta $, for $ 0 \leq \theta < 2\pi $, where $ 2\sin^2\theta = -\sin\theta $?

(A) $ 0, \pi, \frac{\pi}{6}, $ and $ \frac{5\pi}{6} $
(B) $ 0, \pi, \frac{7\pi}{6}, $ and $ \frac{11\pi}{6} $
(C) $ \frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{3}, $ and $ \frac{5\pi}{3} $
(D) $ \frac{\pi}{2}, \frac{3\pi}{2}, \frac{2\pi}{3}, $ and $ \frac{4\pi}{3} $

▶️ Answer/Explanation

Detailed solution

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)

Question

The function $ f $ is given by $ f(x) = \sin(2.25x + 0.2) $. The function $ g $ is given by $ g(x) = f(x) + 0.5 $. What are the zeros of $ g $ on the interval $ 0 \leq x \leq \pi $?

(A) 1.085 and 2.481
(B) 1.307 and 2.704
(C) 1.540 and 2.471
(D) 0.144, 1.075, and 2.936

▶️ Answer/Explanation

Detailed solution

$ 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)

Question

What is the angle of $\tan \theta = 3$ if $\cos \theta < 0$ and $0 \leq \theta < 2\pi$?

a. $\theta = 1.25$
b. $\theta = 181.25$
c. $\theta = 4.39$
d. $\theta = 7.53$

▶️ Answer/Explanation

Detailed solution

First, find the reference angle using $\alpha = \arctan(3) \approx 1.25$ radians.
Given $\tan \theta = 3$ (positive) and $\cos \theta < 0$ (negative), the angle must be in Quadrant III.

In Quadrant III, the formula for the angle is $\theta = \pi + \alpha$.
Substitute the values: $\theta = 3.14159 + 1.25$.
The resulting calculation is $\theta \approx 4.39$ radians.
This value falls within the required range of $0 \leq \theta < 2\pi$.
Therefore, the correct option is c.

Question

Find all the solutions in the interval $[0, 2\pi)$: $\sin(2x) + \sin(x) = 0$.

(A) $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{2\pi}{3}, \frac{4\pi}{3}$

(B) $0, \frac{\pi}{3}, \frac{\pi}{6}, \frac{5\pi}{3}$

(D) $0, \frac{\pi}{3}$

▶️ Answer/Explanation

Detailed solution

Use the double-angle identity $\sin(2x) = 2\sin(x)\cos(x)$ to rewrite the equation:
$2\sin(x)\cos(x) + \sin(x) = 0$
Factor out $\sin(x)$:
$\sin(x)(2\cos(x) + 1) = 0$
This gives two cases: $\sin(x) = 0$ or $2\cos(x) + 1 = 0$.
Case 1: $\sin(x) = 0 \implies x = 0, \pi$ (in the given interval).
Case 2: $\cos(x) = -\frac{1}{2} \implies x = \frac{2\pi}{3}, \frac{4\pi}{3}$ (cosine is negative in Q2 and Q3).
The complete solution set is $\{0, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}\}$.
Note: Option (C) is the closest match, likely containing a typo where $\pi$ was misprinted as $\frac{\pi}{3}$.

Question

The function $ g $ is defined by $ g(x) = \sin \left( x + \frac{\pi}{3} \right) $. The solutions to which of the following equations on the interval $ 0 \leq x \leq 2\pi $ are the solutions to $ g(x) = 1 $

(A) $ \sin x + \sqrt{3} \cos x = 2 $

(B) $ \sin x – \sqrt{3} \cos x = 2 $

(D) $ \sqrt{3} \sin x – \cos x = 2 $

▶️ Answer/Explanation

Detailed solution

We are looking for an equation equivalent to $ g(x) = 1 $, where $ g(x) = \sin \left( x + \frac{\pi}{3} \right) $.
Using the sum identity for sine, $ \sin(A + B) = \sin A \cos B + \cos A \sin B $, we expand the left side:
$ \sin x \cos \left( \frac{\pi}{3} \right) + \cos x \sin \left( \frac{\pi}{3} \right) = 1 $
Substitute the known trigonometric values $ \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} $ and $ \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} $:
$ \sin x \left( \frac{1}{2} \right) + \cos x \left( \frac{\sqrt{3}}{2} \right) = 1 $
Multiplying the entire equation by 2 to eliminate the denominators gives:
$ \sin x + \sqrt{3} \cos x = 2 $
Therefore, the correct option is (A).

Question

The function $ g $ is given by $ g(x) = 2 \cos(x) $. What are all solutions for when $ g(x) = \sqrt{3} $?

(A) $ x = \frac{\pi}{6} + 2\pi k $ and $ x = \frac{5\pi}{6} + 2\pi k $, where $ k $ is any integer.

(B) $ x = \pm \frac{\pi}{6} + 2\pi k $, where $ k $ is any integer.

(D) $ x = \pm \frac{\pi}{3} + 2\pi k $, where $ k $ is any integer.

▶️ Answer/Explanation

Detailed solution

We begin by setting the function $ g(x) $ equal to $ \sqrt{3} $: $ 2 \cos(x) = \sqrt{3} $.

Divide both sides by $ 2 $ to isolate the cosine function: $ \cos(x) = \frac{\sqrt{3}}{2} $.

We identify the reference angle in the first quadrant where $ \cos(x) = \frac{\sqrt{3}}{2} $, which is $ \frac{\pi}{6} $.

Since cosine is positive in the first and fourth quadrants, the solutions in one period are $ x = \frac{\pi}{6} $ and $ x = -\frac{\pi}{6} $.

The period of cosine is $ 2\pi $, so we add $ 2\pi k $ (where $ k $ is an integer) to these base solutions.

This results in the general solutions: $ x = \frac{\pi}{6} + 2\pi k $ and $ x = -\frac{\pi}{6} + 2\pi k $.

Combining these expressions concisely, we get $ x = \pm \frac{\pi}{6} + 2\pi k $.

Comparing this result with the options, it matches option (B).

Question

Consider the sinusoidal functions $f$ and $g$, defined by $f(x) = 3\cos(x + 2.78)$ and $g(x) = \sin(x – 0.25)$. Which of the following describes a zero of the function $h$ defined by $h(x) = f(x) – g(x)$ on the interval $[0, \pi]$?

(A) $2.242$

(B) $1.932$

(D) $0.250$

▶️ Answer/Explanation

Detailed solution

A zero of the function $h(x)$ occurs where $h(x) = 0$.
This means we must find $x$ such that $f(x) – g(x) = 0$, or $f(x) = g(x)$.
Substitute the functions: $3\cos(x + 2.78) = \sin(x – 0.25)$.
Using a graphing utility, plot $y = 3\cos(x + 2.78) – \sin(x – 0.25)$ on $[0, \pi]$.
Observe the point where the graph crosses the $x$-axis within the interval.
The intersection occurs at $x \approx 2.242$.
Thus, the zero of the function in the given interval is $2.242$.
The correct option is (A).

Question

The function $k$ is given by $k(\theta) = 2 \sin \theta$. What are all values of $\theta$, for $0 \le \theta < 2\pi$, where $k(\theta) = -1$?

(A) $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$

(B) $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$

(D) $\theta = \frac{7\pi}{6}$ and $\theta = \frac{11\pi}{6}$

▶️ Answer/Explanation

Detailed solution

Set the function equal to the target value: $2 \sin \theta = -1$.
Isolate the sine function by dividing both sides by $2$ to get $\sin \theta = -\frac{1}{2}$.
Identify the reference angle where $\sin \theta = \frac{1}{2}$, which is $\theta_{ref} = \frac{\pi}{6}$.
Since the sine value is negative, $\theta$ must be in Quadrant III or Quadrant IV.
In Quadrant III: $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.
In Quadrant IV: $\theta = 2\pi – \frac{\pi}{6} = \frac{11\pi}{6}$.
Therefore, the correct option is (D).

Question

The function $ k $ is given by $ k(\theta) = 2\sin \theta $. What are all values of $ \theta $, for $ 0 \le \theta < 2\pi $, where $ k(\theta) = -1 $?

(A) $ \theta = \frac{\pi}{6} $ and $ \theta = \frac{5\pi}{6} $

(B) $ \theta = \frac{\pi}{3} $ and $ \theta = \frac{5\pi}{3} $

(D) $ \theta = \frac{7\pi}{6} $ and $ \theta = \frac{11\pi}{6} $

▶️ Answer/Explanation

Detailed solution

Set the function equal to the target value: $ 2\sin \theta = -1 $.
Isolate the sine term by dividing both sides by 2: $ \sin \theta = -\frac{1}{2} $.
Identify the reference angle: $ \sin \frac{\pi}{6} = \frac{1}{2} $, so the reference angle is $ \frac{\pi}{6} $.
Since $ \sin \theta $ is negative, $ \theta $ must lie in Quadrant III or Quadrant IV.
For Quadrant III: $ \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} $.
For Quadrant IV: $ \theta = 2\pi – \frac{\pi}{6} = \frac{11\pi}{6} $.
Therefore, the values are $ \frac{7\pi}{6} $ and $ \frac{11\pi}{6} $, which corresponds to option (D).

Question

The functions $f$ and $g$ are given by $f(\theta) = \sin\theta$ and $g(\theta) = \sin(\theta + \frac{\pi}{2})$. On the interval $0 \leq \theta \leq 2\pi$, how many solutions exist to $f(\theta) = g(\theta)$?

(A) Zero
(B) One
(C) Two
(D) Infinitely many

▶️ Answer/Explanation

Detailed solution

Using the identity $\sin(\theta + \frac{\pi}{2}) = \cos\theta$, we set $f(\theta) = g(\theta)$ as $\sin\theta = \cos\theta$.
Dividing by $\cos\theta$ (where $\cos\theta \neq 0$), we get $\tan\theta = 1$.
In the interval $[0, 2\pi]$, the tangent function equals $1$ at two specific points.
The first solution occurs in Quadrant I at $\theta = \frac{\pi}{4}$.
The second solution occurs in Quadrant III at $\theta = \frac{5\pi}{4}$.
Therefore, there are exactly two solutions within the given interval.
The correct option is (C).

Question

A second hand on a clock rotates clockwise at a constant speed around the clock. However, Charlie bought a new clock that is dysfunctional, where the second hand completes a cycle every (30) seconds. The clock is (24) inches in diameter and is (5) feet above the ground. The distance between the point at which the second hand is pointing to and the ground is periodically increasing and decreasing. The second hand starts at (t = 0) pointing straight up towards the number (12).

(A) This situation can be represented with a sinusoidal function in the form $h(t) = a \cos( b(t + c) ) + d$, where $t$ is the number of seconds after starting the clock and $h(t)$ is the height of the end of the second hand in inches. Find the function.

(B) When does the arrow of the second hand become $72 – 6\sqrt{2}$ inches above the ground on the interval $0 < t \leq 30$?

(C) i. Between $0 \leq t \leq 30$, does the graph of $h(t)$ have a point of inflection? If so, identify all the points of inflection between the time intervals. If it does not, explain why using your knowledge of a cosine function and its continuity.

ii. Explain the meaning of these points in context of the problem using the definition of the rate of change and the rate the rate is changing.

Most-appropriate topic codes (CED):

• TOPIC 3.7: Sinusoidal Function Context and Data Modeling — part (A)
• TOPIC 3.10: Trigonometric Equations and Inequalities — part (B)
• TOPIC 1.1: Change in Tandem — part (C)

▶️ Answer/Explanation

Detailed solution

(A)
First, determine the physical parameters of the clock to find the sinusoidal constants.
The diameter is (24) inches, so the radius is (12) inches. This corresponds to the amplitude, (a = 12).
The center of the clock is (5) feet above the ground, which converts to (60) inches. This is the midline, (d = 60).
The second hand completes a cycle every (30) seconds. The period is (30), so the frequency coefficient is (b = \dfrac{2\pi}{30} = \dfrac{\pi}{15}).
At (t = 0), the hand is pointing straight up (maximum height). Since the cosine function starts at a maximum, there is no phase shift, so (c = 0).
Thus, the function is:
(h(t) = 12 \cos\left( \dfrac{\pi}{15} t \right) + 60)

(B)
Set $h(t)$ equal to the given height and solve for $t$:
$12 \cos\left( \dfrac{\pi}{15} t \right) + 60 = 72 – 6\sqrt{2}$
$12 \cos\left( \dfrac{\pi}{15} t \right) = 12 – 6\sqrt{2}$
$\cos\left( \dfrac{\pi}{15} t \right) = 1 – \dfrac{\sqrt{2}}{2} = \dfrac{2 – \sqrt{2}}{2}$
Let $\theta = \dfrac{\pi}{15} t$. We need to solve $\cos \theta = \dfrac{2 – \sqrt{2}}{2}$.
$\theta = \pm \arccos\left( \dfrac{2 – \sqrt{2}}{2} \right) + 2\pi k$.
For the interval $0 < t \leq 30$, we look for solutions in $(0, 2\pi]$.
The two solutions for $\theta$ are $\theta_1 = \arccos\left( \dfrac{2 – \sqrt{2}}{2} \right)$ and $\theta_2 = 2\pi – \theta_1$.
Converting back to time $t = \dfrac{15}{\pi} \theta$:
$t_1 = \dfrac{15}{\pi} \arccos\left( \dfrac{2 – \sqrt{2}}{2} \right)$ and $t_2 = 30 – t_1$.

(C) i.
Yes, the graph of $h(t)$ has points of inflection. The function $h(t)$ is a smooth, continuous cosine wave.
Points of inflection on a sinusoidal graph occur where the graph crosses its midline (concavity changes from up to down or vice versa). This happens when:
$\cos\left( \dfrac{\pi}{15} t \right) = 0$
$\dfrac{\pi}{15} t = \dfrac{\pi}{2} + k\pi$
$t = 15\left( \dfrac{1}{2} + k \right) = 7.5 + 15k$
For the interval $0 \leq t \leq 30$:
If $k=0$, $t = 7.5$.
If $k=1$, $t = 22.5$.
The points of inflection are at $t = 7.5$ seconds and $t = 22.5$ seconds.

(C) ii.
In this context:
The rate of change represents the vertical velocity of the tip of the second hand.
The rate the rate is changing represents the vertical acceleration.
A point of inflection occurs where the concavity changes, which means the vertical acceleration is transitioning from positive to negative (or vice versa) and is instantaneously zero. Physically, these are the moments when the tip of the second hand is at the same height as the center of the clock (midline). At these specific points, the vertical speed (rate of change) is at its absolute maximum magnitude.

Question

The functions $ g $ and $ h $ are given by:
$$ g(x) = 2 \log a – 3 \log b + \frac{\log a}{2} + \log (100 c) $$
$$ h(x) = \cot x \sec^{2} x – \tan x $$

(A) i. Rewrite $ g(x) $ as a single logarithm with base 10 without negative exponents in any part of the expression. Your result should be simplified in the form $ \log(\text{expression}) $.
ii. Rewrite $ h(x) $ as an expression in which $ \cot x $ appears once and no other trigonometric functions are involved.

(B) The functions $ j $ and $ k $ are given by:
$$ j(x) = 3 \cot^{2} \theta + 3 \csc \theta $$
$$ k(x) = 7 e^{-2x} + 9 $$

i. Solve $ j(x) = -3 $ for values of $ x $ in the interval $ [0, 2\pi) $.
ii. Solve $ k(x) = 86 $ for all values of $ x $ in the domain of $ k $.

(C) Two functions $ a $ and $ b $ are given by:
$$ a(x) = 3 – 3 \csc 3\theta $$
$$ b(x) = -1 + \csc 3\theta $$

Find all the input values in the domain of $ a $ and $ b $ where $ a(x) $ and $ b(x) $ intersect.

Most-appropriate topic codes (CED):

• TOPIC 2.12: Logarithmic Function Manipulation — part (A) i
• TOPIC 3.11: Equivalent Representations of Trigonometric Functions — part (A) ii
• TOPIC 3.12: Trigonometric Equation Solving — part (B) i
• TOPIC 2.13: Exponential and Logarithmic Equations and Inequalities — part (B) ii
• TOPIC 3.12: Trigonometric Equation Solving — part (C)

▶️ Answer/Explanation

Detailed solution

(A) i. Logarithmic Simplification
Start with the given expression:
$ g(x) = 2 \log a + \frac{1}{2} \log a – 3 \log b + \log (100 c) $

Combine the $ \log a $ terms ($ 2 + 0.5 = 2.5 $):
$ \frac{5}{2} \log a – 3 \log b + \log (100 c) $

Use power rules to move coefficients inside the logarithms:
$ \log (a^{5/2}) – \log (b^3) + \log (100 c) $

Combine using product and quotient rules:
$ \log \left( \frac{100 c \cdot a^{5/2}}{b^3} \right) $

Final Answer: $ \log \left( \frac{100 a^{5/2} c}{b^3} \right) $

(A) ii. Trigonometric Simplification
Start with: $ h(x) = \cot x \sec^2 x – \tan x $

Convert to sine and cosine:
$ \left(\frac{\cos x}{\sin x}\right) \cdot \left(\frac{1}{\cos^2 x}\right) – \frac{\sin x}{\cos x} $

Simplify the first term:
$ \frac{1}{\sin x \cos x} – \frac{\sin x}{\cos x} $

Find a common denominator ($ \sin x \cos x $) and simplify:
$ \frac{1 – \sin^2 x}{\sin x \cos x} = \frac{\cos^2 x}{\sin x \cos x} $

Cancel terms:
$ \frac{\cos x}{\sin x} $

Final Answer: $ \cot x $

(B) i. Trigonometric Equation
Set $ j(x) = -3 $:
$ 3 \cot^2 \theta + 3 \csc \theta = -3 $

Divide by 3 and rearrange:
$ \cot^2 \theta + \csc \theta + 1 = 0 $

Substitute identity $ \cot^2 \theta = \csc^2 \theta – 1 $:
$ (\csc^2 \theta – 1) + \csc \theta + 1 = 0 $
$ \csc^2 \theta + \csc \theta = 0 $

Factor:
$ \csc \theta (\csc \theta + 1) = 0 $

Solve $ \csc \theta = -1 $ (since $ \csc \theta \neq 0 $):
$ \sin \theta = -1 $

[Image of unit circle sine values]

In the interval $ [0, 2\pi) $:
Final Answer: $ \theta = \frac{3\pi}{2} $

(B) ii. Exponential Equation
Set $ k(x) = 86 $:
$ 7 e^{-2x} + 9 = 86 $

Subtract 9 and divide by 7:
$ 7 e^{-2x} = 77 \implies e^{-2x} = 11 $

Take the natural log:
$ -2x = \ln 11 $

Final Answer: $ x = -\frac{1}{2} \ln 11 $

(C) Intersection
Set $ a(x) = b(x) $:
$ 3 – 3 \csc 3\theta = -1 + \csc 3\theta $

Rearrange:
$ 4 = 4 \csc 3\theta \implies \csc 3\theta = 1 $

So, $ \sin 3\theta = 1 $. The sine function equals 1 at $ \frac{\pi}{2} $ plus full rotations:
$ 3\theta = \frac{\pi}{2} + 2k\pi $

Final Answer: $ \theta = \frac{\pi}{6} + \frac{2k\pi}{3} $, where $ k \in \mathbb{Z} $