Difference between revisions of "2021 AMC 12A Problems/Problem 19"
MRENTHUSIASM (talk | contribs) m (→Solution 3 (Graphs and Analyses)) |
MRENTHUSIASM (talk | contribs) (Condensed Sol 2 a bit.) |
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==Solution 2 (Cofunction Identity)== | ==Solution 2 (Cofunction Identity)== | ||
− | By the | + | By the Cofunction Identity <math>\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),</math> we simplify the given equation: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \sin \left( \frac{\pi}2 \cos x\right) &= | + | \sin \left( \frac{\pi}2 \cos x\right) &= \sin \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) \\ |
− | + | \frac{\pi}2 \cos x &= \frac{\pi}2 - \frac{\pi}2 \sin x + 2n\pi | |
− | |||
− | \frac{\pi}2 | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | for some integer <math>n.</math> We | + | for some integer <math>n.</math> |
− | <cmath>\ | + | |
− | + | We rearrange and simplify: <cmath>\sin x + \cos x = 1 + 4n.</cmath> | |
− | + | By rough constraints, we know that <math>-2 < \sin x + \cos x < 2,</math> from which <math>-2 < 1 - 4n < 2.</math> The only possibility is <math>n=0,</math> so we get | |
− | |||
− | By rough constraints, we know that <math>-2 < \sin x + \cos x < 2,</math> from which <math>-2 < 1 - 4n < 2.</math> The only possibility is <math>n=0 | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \sin x + \cos x &= 1 | + | \sin x + \cos x &= 1 && (*) \\ |
\sin^2 x + \cos^2 x + 2\sin x \cos x &= 1 \\ | \sin^2 x + \cos^2 x + 2\sin x \cos x &= 1 \\ | ||
2\sin x \cos x &= 0 \\ | 2\sin x \cos x &= 0 \\ | ||
\sin(2x) &= 0 \\ | \sin(2x) &= 0 \\ | ||
2x &= k\pi \\ | 2x &= k\pi \\ | ||
− | x &= \frac{k\pi}{2} | + | x &= \frac{k\pi}{2} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
for some integer <math>k.</math> | for some integer <math>k.</math> | ||
− | The <i><b>possible</b></i> solutions in <math>[0,\pi]</math> are <math>x=0,\frac{\pi}{2},\pi | + | The <i><b>possible</b></i> solutions in <math>[0,\pi]</math> are <math>x=0,\frac{\pi}{2},\pi.</math> However, <math>x=\pi</math> is an extraneous solution by squaring <math>(*).</math> Therefore, the answer is <math>\boxed{\textbf{(C) }2}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 22:57, 16 September 2021
Contents
Problem
How many solutions does the equation have in the closed interval
?
Solution 1 (Inverse Trigonometric Functions)
The ranges of and
are both
which is included in the range of
so we can use it with no issues.
This only happens at
on the interval
because one of
and
must be
and the other
Therefore, the answer is
~Tucker
Solution 2 (Cofunction Identity)
By the Cofunction Identity we simplify the given equation:
for some integer
We rearrange and simplify:
By rough constraints, we know that
from which
The only possibility is
so we get
for some integer
The possible solutions in are
However,
is an extraneous solution by squaring
Therefore, the answer is
~MRENTHUSIASM
Solution 3 (Graphs and Analyses)
Let and
This problem is equivalent to counting the intersections of the graphs of
and
in the closed interval
We make a table of values, as shown below:
For the graphs of
and
we will analyze their increasing/decreasing behaviors in
- The graph of
in
(from left to right) has the same behavior as the graph of
in
(from right to left): The output is from
to
(from left to right), inclusive, and strictly decreasing.
- The graph of
in
(from left to right) has two parts:
- The graph of
in
has the same behavior as the graph of
in
(from left to right): The output is from
to
(from left to right), inclusive, and strictly decreasing.
- The graph of
in
has the same behavior as the graph of
in
(from right to left): The output is from
to
(from left to right), inclusive, and strictly increasing.
If then
and
So, their graphs do not intersect.
If then
Clearly, the graphs intersect at
and
(at points
and
respectively), but we will determine whether they are the only points of intersection:
Let and
It follows that
Since
we know that
by the cofunction identity:
From the last block of equations in Solution 2, we conclude that
and
are the only points of intersection. So, the answer is
~MRENTHUSIASM (credit given to TheAMCHub)
Remark
The graphs of (in red) and
(in blue) are shown below.
Graph in Desmos: https://www.desmos.com/calculator/brjh3gybcc
~MRENTHUSIASM
Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry)
~ pi_is_3.14
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.