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).

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

AP Precalculus -3.10 Trigonometric Equations and Inequalities- MCQ Exam Style Questions - Effective Fall 2023