2022 AMC 12A Problem 17

Question

Suppose

a

is a real number such that the equation

a\cdot(\sin x+\sin(2x))=\sin(3x)

has more than one solution in the interval

(0,\pi).

The set of all such

a

can be written in the form

(p,q)\cup(q,r),

where

p,

q,

and

r

are real numbers with

p\lt q\lt r.

What is

p+q+r?

-4

-1

0

1

4

Use

\sin 2x=2\sin x\cos x

and

\sin 3x=\sin x(4\cos^2 x-1)

After dividing by

\sin x,

note

x=\dfrac{2\pi}{3}

(where

\cos x=-\tfrac12

) is always a solution

Since

\sin x\ne0

on

(0,\pi),

divide by

\sin x:

\begin{gathered} a(1+2\cos x)=4\cos^2 x-1 \\ =(2\cos x-1)(2\cos x+1). \end{gathered}

When

\cos x=-\tfrac12

(that is,

x=\tfrac{2\pi}{3}

) both sides vanish, so this is a solution for every

a.

Otherwise we may cancel

1+2\cos x

to get

a=2\cos x-1,

i.e.

\cos x=\dfrac{a+1}{2}.

This yields a second solution in

(0,\pi)

exactly when

-1\lt\dfrac{a+1}{2}\lt1,

that is

a\in(-3,1),

and it is distinct from

x=\tfrac{2\pi}{3}

unless

a=-2.

So more than one solution occurs for

a\in(-3,-2)\cup(-2,1),

giving

p+q+r=-3-2+1=-4.

Thus, the correct answer is A.

Problem 17 in other years