AMC 10 · 2009 · #24

Grade 11 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

trigonometric-ratiosfunction-compositiondomain-restrictionperiodic-function easier-related-problemsystematic-enumerationcasework ↑ Prerequisites: trigonometric-ratiosfunction-composition

📏 Long solution 💡 4 insights

Problem

Two inverse trigonometric expressions in one variable are set equal on a closed interval. Count the solutions.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Solve an Easier Related Problem

The equation as written involves a composition that is piecewise, so tool #9 (Solve an Easier Related Problem) is primary: replace it with the plain equation sin 6x = sin x carrying an interval condition, and solve that instead. The replacement is the whole problem, so it must be done as an equivalence and not as a one-way consequence. Applying sine to both sides is legal and loses nothing, but it also proves nothing on its own: it produces candidates that need not be solutions, and there really are such impostors here. What restores the missing direction is tool #14 (Extreme Principle), applied to the range of sin⁻¹ rather than to the algebra: the left side can never exceed π/2, so no x past π/2 can work. That single bound is not a tidy-up at the end; it is half the answer, and skipping it lands you on a wrong choice that is sitting in the list waiting. Tool #1 (Draw a Diagram) supplies the complete description of when two angles share a sine, read off the unit circle, so the case list is provably complete rather than remembered. Tools #13 and #2 then finish mechanically, and tool #6 checks every survivor against the original equation.

1STEP 1

The right side is exactly x

One side simplifies to the variable itself.

cos⁻¹(cos x) = x for every x ∈ [0, π], so the equation becomes sin⁻¹(sin 6x) = x

2STEP 2

Half the interval dies at once

The other's range kills half the interval.

-π/2 ≤ sin⁻¹(t) ≤ π/2 ⟹ x = sin⁻¹(sin 6x) ≤ π/2, so every solution lies in [0, π/2]

3STEP 3

Trade the inverse for an equivalence

On what is left the inverse can be traded for an equivalence.

For 0 ≤ x ≤ π/2: sin⁻¹(sin 6x) = x ⇔ sin 6x = sin x

4STEP 4

When two angles share a sine

Two sines being equal gives two families.

sin A = sin B ⇔ A = B + 2kπ or A = π - B + 2kπ, k ∈ Z

5STEP 5

Two ladders of candidates

Each family is an arithmetic ladder of candidates.

6x = x + 2kπ → x = 2kπ/5, 6x = π - x + 2kπ → x = ((2k+1)π)/7

6STEP 6

Keep only the rungs below pi over 2

Keeping only the rungs inside the range leaves a few.

0 ≤ 2kπ/5 ≤ π/2 ⇔ k ∈ {0, 1}, 0 ≤ ((2k+1)π)/7 ≤ π/2 ⇔ k ∈ {0, 1}, 2π/5 = 14π/35 ＜ 15π/35 = 3π/7

7STEP 7

Verify each survivor and count

Verifying each gives 4, choice (B).

sin⁻¹(sin 18π/7) = sin⁻¹(sin 4π/7) = sin⁻¹(sin 3π/7) = 3π/7 = cos⁻¹(cos 3π/7), 4 values → (B)

Answer

The best check is to find out where the wrong choices come from, because one of them is manufactured by exactly the step this solution had to prove. Solve sin 6x = sin x on all of [0, π] without the range cap: the first family gives 0, 2π/5, 4π/5 and the second gives π/7, 3π/7, 5π/7, π, which is 7 values, sitting in the list as choice (E). The cap x ≤ π/2 deletes the three largest, leaving 4. So the two numbers that matter are related in a way that confirms both: 7 candidates, 3 impostors, 4 solutions. Each impostor can be exposed by hand. At x = 4π/5 the left side is π/5; at x = 5π/7, 6x = 30π/7 = 4π + 2π/7 so the left side is 2π/7; at x = π the left side is sin⁻¹(sin 6π) = 0. None equals its own x, and each failure is the same failure: the true value of sin⁻¹ is the reflected angle, not the one you started with. The boundary deserves its own look, since a fifth solution could hide there: at x = π/2 the left side is sin⁻¹(sin 3π) = 0 ≠ π/2, so the cut point is not a solution and nothing was gained or lost by including it. A numerical spot check on the least obvious root: x = 3π/7 ≈ 1.3464, 6x ≈ 8.0784, sin 6x ≈ 0.9749, and sin⁻¹(0.9749) ≈ 1.3464, which matches x.

💡Key takeaway

An inverse trig function only ever hands back an angle from its own narrow branch, so before matching sines, first ask which values of x the answer is even allowed to be.

The right side is exactly x
Half the interval dies at once
Trade the inverse for an equivalence
When two angles share a sine
Two ladders of candidates
Keep only the rungs below pi over 2
Verify each survivor and count