Difference between revisions of "2021 AMC 12A Problems/Problem 19"
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==Solution== | ==Solution== | ||
| − | {{ | + | <math>\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)</math> |
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| + | The interval is <math>[0,\pi]</math>, so both <math>\arccos</math> and <math>\arcsin</math> will lie within the range. | ||
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| + | <math>\frac{\pi}2 \cos x=\arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right)</math> | ||
| + | |||
| + | <math>\frac{\pi}2 \cos x=\frac{\pi}2 - \frac{\pi}2 \sin x</math> | ||
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| + | <math>\cos x = 1 - \sin x</math> | ||
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| + | This only happens at <math>x = 0, \frac{\pi}2</math> on the interval <math>[0,\pi]</math>, because either <math>\cos x = 0</math> and <math>\sin x = 1</math> or <math>\cos x = 1</math> and <math>\sin x = 0</math>. | ||
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| + | ~Tucker | ||
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==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=18|num-a=20}} | {{AMC12 box|year=2021|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:57, 11 February 2021
Problem
How many solutions does the equation 
have in the closed interval ![$[0,\pi]$](http://latex.artofproblemsolving.com/e/c/e/ece27ba559bd93cdf2416aa0f88fc8d7d85e5791.png)
?
Solution
The interval is , so both 
and 
will lie within the range.
This only happens at 
on the interval , because either 
and 
or 
and 
.
~Tucker
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
