AMC 10 · 2017 · #15

Grade 11 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

trigonometric-ratiosdomain-restrictionperiodic-function sign-analysisidentify-subproblems ↑ Prerequisites: trigonometric-ratiosdomain-restriction

📏 Medium solution 💡 3 insights

Problem

A sum of three trigonometric terms must first reach zero. Find the interval holding that point.

Pick an answer.

(A)

(0,1)

(B)

(1, 2)

(C)

(2, 3)

(D)

(3, 4)

(E)

(4,5)

How to solve

Strategy Identify Subproblems

Solving sin x + 2cos x + 3tan x = 0 exactly is hopeless, and the question does not ask for the root — only for the interval holding it. Tool #7 (Identify Subproblems) does the heavy lifting: tan x blows up at x = π/2 + kπ, so cut the positive number line at those points and study f on one unbroken piece at a time. Tool #15 (Organize Information in More Ways) rewrites f using tan x = (sin x)/(cos x) so that its sign can be read off directly instead of estimated. Tool #6 (Guess and Check) supplies one well-chosen test point, x = 5π/4, where all three trig values are exact. Tool #3 (Eliminate Possibilities) then knocks out the early intervals, which is what "smallest" demands.

1STEP 1

Find where f has no value

One term is undefined, cutting the line into pieces.

f undefined at x = π/2 + kπ → pieces (0, π/2), (π/2, 3π/2), (3π/2, 5)

2STEP 2

Everything is positive before pi/2

Everything is positive on the first piece.

0 ＜ x ＜ π/2 → sin x ＞ 0, cos x ＞ 0, tan x ＞ 0 → f(x) ＞ 0

3STEP 3

Rewrite f to read its sign

Rewriting makes the next piece's sign readable.

f(x) = sin x(1 + 3/(cos x)) + 2cos x, π/2 ＜ x ≤ π → f(x) ＜ 0

4STEP 4

Nothing crosses zero below pi

So no crossing happens early.

f ＞ 0 on (0, π/2), f undefined at π/2, f ＜ 0 on (π/2, π]

5STEP 5

Test the exact point 5pi/4

One exact point tests positive again.

f(5π/4) = -√(2)/2 + 2(-√(2)/2) + 3(1) = 3 - 3√(2)/2 ＞ 0

6STEP 6

A sign change traps the root

That sign change traps the root, choice (D).

f(π) = -2 ＜ 0 ＜ f(5π/4) → ∃ x₀ ∈ (π, 5π/4) ⊂ (3,4) with f(x₀) = 0 → (D)

Answer

(3, 4)

Tighten the bracket to see the root is well inside (3,4), not hugging an endpoint: f(3.7) ≈ -0.35 and f(3.8) ≈ 0.13, so the smallest positive zero is near x ≈ 3.78. That is comfortably inside (3,4), confirming (D). The tempting wrong pick is (B): f really does swing from huge positive values to huge negative values across x = π/2 ≈ 1.571, which lies inside (1,2). But f has no value at π/2, so that swing is an asymptote and not a root — checking the domain first is what keeps (B) off the table.

💡Key takeaway

Cut the line where tangent blows up, check the sign on each piece, and the first place the sign truly flips is where the root hides.

Find where f has no value
Everything is positive before pi/2
Rewrite f to read its sign
Nothing crosses zero below pi
Test the exact point 5pi/4
A sign change traps the root