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Difference between revisions of "2022 AMC 12A Problems/Problem 17"

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

Revision as of 00:29, 14 November 2022

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 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

See Also

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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