# 2022 AMC 12A Problems/Problem 17

## Problem

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 $a$ 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 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

## Solution 3

Use the sum to product formula to obtain $2a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{3x}$. Use the double angle formula on the RHS to obtain $a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{\frac{3x}{2}}\cos{\frac{3x}{2}}$. From here, it is obvious that $x=\frac{2\pi}{3}$ is always a solution, and thus we divide by $\sin{\frac{3x}{2}}$ to get $$a\cdot\cos{\frac{x}{2}}=\cos{\frac{3x}{2}}$$ We wish to find all $a$ such that there is at least one more solution to this equation distinct from $x=\frac{2\pi}{3}$. Letting $y=\cos{\frac{x}{2}}$, and noting that $\cos{\frac{3x}{2}}=4y^3-3y$, we can rearrange our equation to $4y^3=y(3+a)$ The smallest value $x$ where $y=0$ is $\pi$, which is not in our domain so we divide by $y$ to obtain $4y^2=a+3$. By the trivial inequality, $a+3\ge{0}$. Furthermore, $y\neq{0}$, so $a+3>0$. Also, if $a=-2$, then the solution to this equation would be shared with $x=\frac{2\pi}{3}$, so there would only be one distinct solution. Finally, because $y\le{1}$ due to the restrictions of a sine wave, and that $y\neq{1}$ due to the restrictions on $x$, we have $-3 with $a\neq{-2}$. Thus, $p=-3,q=-2, r=1$, so our final answer is $-3+(-2)+1=\boxed{\textbf{(A) } {-}4}$.

~sigma

## Video Solution 1 (Quick and Simple)

~Education, the Study of Everything

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