Difference between revisions of "2021 AMC 12A Problems/Problem 19"
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<math>\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)</math> | <math>\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)</math> | ||
− | The interval is <math>[0,\pi]</math>, which is included in the range | + | The interval is <math>[0,\pi]</math>, which is included in the range <math>\arccos</math>, so we can use it with no issues. |
− | <math>\frac{\pi}2 \ | + | <math>\frac{\pi}2 \sin x=\arccos \left( \sin \left( \frac{\pi}2 \cos x\right)\right)</math> |
− | <math>\frac{\pi}2 \ | + | <math>\frac{\pi}2 \sin x=\frac{\pi}2 - \frac{\pi}2 \cos x</math> |
− | <math>\ | + | <math>\sin x = 1 - \cos x</math> |
<math>\cos x + \sin x = 1</math> | <math>\cos x + \sin x = 1</math> | ||
− | This only happens at <math>x = 0, \frac{\pi}2</math> on the interval <math>[0,\pi]</math>, because one of <math>\sin</math> and <math>\cos</math> must be <math>1</math> and the other <math>0</math>. Therefore, the answer is <math>\boxed{C | + | This only happens at <math>x = 0, \frac{\pi}2</math> on the interval <math>[0,\pi]</math>, because one of <math>\sin</math> and <math>\cos</math> must be <math>1</math> and the other <math>0</math>. Therefore, the answer is <math>\boxed{C) 2}</math> |
~Tucker | ~Tucker | ||
− | |||
== Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry) == | == Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry) == |
Revision as of 13:08, 12 February 2021
Contents
[hide]Problem
How many solutions does the equation have in the closed interval ?
Solution
The interval is , which is included in the range , 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
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.