Difference between revisions of "2022 AMC 12A Problems/Problem 17"
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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)</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>. |
Applying the quadratic formula to solve for <math>\cos{x}</math>, we get <cmath>\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}</cmath> and expanding the terms under the radical, we get <cmath>\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}</cmath> | Applying the quadratic formula to solve for <math>\cos{x}</math>, we get <cmath>\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}</cmath> and expanding the terms under the radical, we get <cmath>\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}</cmath> |
Revision as of 03:28, 12 November 2022
Problem
Supppose is a real number such that the equation
has more than one solution in the interval
. The set of all such
that can be written
in the form
where
and
are real numbers with
. What is
?
Solution
We are given that
Using the sine double angle formula combine with the fact that , which can be derived using sine angle addition with
, we have
Since
as it is on the open interval
, we can divide out
from both sides, leaving us with
Now, distributing
and rearranging, we achieve the equation
which is a quadratic in
.
Applying the quadratic formula to solve for , we get
and expanding the terms under the radical, we get
Factoring, since
, we can simplify our expression even further to
Now, solving for our two solutions, and
.
Since yields a solution that is valid for all
, that being
, we must now solve for the case where
yields a valid value.
As ,
, and therefore
, and
.
There is one more case we must consider inside this interval though, the case where , as this would lead to a double root for
, yielding only one valid solution for
. Solving for this case,
.
Therefore, combining this fact with our solution interval, , so the answer is
- DavidHovey