2022 AMC 12A Problems/Problem 17

Problem

Supppose $a$ 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 $a$ that can be written in the form $(p,q) \cup (q,r),$ where $p, q,$ and $r$ are real numbers with $p < q< r$ . What is $p+q+r$ ?

$\textbf{(A) } -4 \qquad \textbf{(B) } -1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4$

Solution

Solution 1:

We are given that $a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}$

Using the sine double angle formula combine with the fact that $\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}+1)$ , which can be derived using sine angle addition with $\sin{(2x + x)}$ , we have $a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)$ Since $\sin{x} \ne 0$ as it is on the open interval $(0, \pi)$ , we can divide out $\sin{x}$ from both sides, leaving us with $a\cdot(1+2\cos{x})=4\cos^2{x}-1$ Now, distributing $a$ and rearranging, we achieve the equation $4\cos^2{x} - 2a\cos{x} - (1+a) = 0$ which is a quadratic in $\cos{x}$ .

Applying the quadratic formula to solve for $\cos{x}$ , we get $\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}$ and expanding the terms under the radical, we get $\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}$ Factoring, since $4a^2+16a+16 = (2a+4)^2$ , we can simplify our expression even further to $\cos{x} =\frac{a\pm(a+2)}{4}$

Now, solving for our two solutions, $\cos{x} = -\frac{1}{2}$ and $\cos{x} = \frac{a+1}{2}$ .

Since $\cos{x} = -\frac{1}{2}$ yields a solution that is valid for all $a$ , that being $x = \frac{2\pi}{3}$ , we must now solve for the case where $\frac{a+1}{2}$ yields a valid value.

As $x\in (0, \pi)$ , $\cos{x}\in (-1, 1)$ , and therefore $\frac{a+1}{2}\in (-1, 1)$ , and $a\in(-3,1)$ .

There is one more case we must consider inside this interval though, the case where $\frac{a+1}{2} = -\frac{1}{2}$ , as this would lead to a double root for $\cos{x}$ , yielding only one valid solution for $x$ . Solving for this case, $a \ne -2$ .

Therefore, combining this fact with our solution interval, $a\in(-3, -2) \cup (-2, 1)$ , so the answer is $-3-2+1 = \boxed{\textbf{(A) -4}}$

- DavidHovey

Solution 2:

We can optimize from the step from $a\cdot(1+2\cos{x})=4\cos^2{x}-1$ in solution 1 by writing

$a = \frac{4\cos^2{x}-1}{1+2\cos{x}} = 2\cos x - 1$

and then get $\cos x = \frac{a+1}{2}.$

Now, solving for our two solutions, $\cos{x} = -\frac{1}{2}$ and $\cos{x} = \frac{a+1}{2}$ .

Since $\cos{x} = -\frac{1}{2}$ yields a solution that is valid for all $a$ , that being $x = \frac{2\pi}{3}$ , we must now solve for the case where $\frac{a+1}{2}$ yields a valid value.

As $x\in (0, \pi)$ , $\cos{x}\in (-1, 1)$ , and therefore $\frac{a+1}{2}\in (-1, 1)$ , and $a\in(-3,1)$ .

There is one more case we must consider inside this interval though, the case where $\frac{a+1}{2} = -\frac{1}{2}$ , as this would lead to a double root for $\cos{x}$ , yielding only one valid solution for $x$ . Solving for this case, $a \ne -2$ .

Therefore, combining this fact with our solution interval, $a\in(-3, -2) \cup (-2, 1)$ , so the answer is $-3-2+1 = \boxed{\textbf{(A) -4}}$

- Dan

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2022 AMC 12A Problems/Problem 17

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