AP Precalculus · Past multiple choice · Question 15 · No calculator

Where a secant graph meets a horizontal line

Set a shifted and stretched secant function equal to a constant, rewrite it as a cosine equation, and solve on one full period.

Lessons this question tests

1 Read the question

The question is College Board's: sample question 15 in AP Precalculus Course and Exam Description, Effective Fall 2026. It starts on page 146 of the PDF. Work it without a calculator, as the exam does.

Open Question 15 (PDF, page 146)

2 Pick your answer

3 Check the explanation

Why the answer is C open once you have picked

Set the two functions equal and solve for $\sec x$ : $1 + 3 \sec x = - 5 \Rightarrow 3 \sec x = - 6 \Rightarrow \sec x = - 2$ Secant is the reciprocal of cosine, so $\cos x = - \frac{1}{2}$ .

Cosine is $\frac{1}{2}$ at the reference angle $\frac{π}{3}$ . It is negative in the second and third quadrants, so on $0 \leq x < 2 π$ : $x = π - \frac{π}{3} = \frac{2 π}{3} and x = π + \frac{π}{3} = \frac{4 π}{3}$ That is choice C.

Check: $\cos \frac{2 π}{3} = - \frac{1}{2}$ , so $\sec \frac{2 π}{3} = - 2$ and $f (\frac{2 π}{3}) = 1 + 3 (- 2) = - 5$ .

Each choice

A. These are the angles where $\cos x = \frac{1}{2}$ , positive. The reference angle is right but the sign was lost, so the angles are in quadrants I and IV instead of II and III.
B. These are the solutions of $\sin x = \frac{1}{2}$ . Two slips combine here: secant was treated as the reciprocal of sine (that is cosecant), and the negative sign was lost.
C (correct). Here $\sec x = - 2$ means $\cos x = - \frac{1}{2}$ , which happens at $\frac{2 π}{3}$ and $\frac{4 π}{3}$ .
D. These solve $\sin x = - \frac{1}{2}$ . The sign is right, but the reciprocal of secant is cosine, not sine.

Question and answer key: College Board. The title and summary are ours, and so is the explanation.