Difference between revisions of "2002 AMC 12P Problems/Problem 10"
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</math> | </math> | ||
− | == Solution == | + | == Solution 1 == |
− | Divide by 2 on both sides to get <cmath>3f_{4}(x)-2f_{6}(x)=f_{2}(x) | + | Divide by 2 on both sides to get <cmath>3f_{4}(x)-2f_{6}(x)=f_{2}(x)</cmath> |
Substituting the definitions of <math>f_{2}(x)</math>, <math>f_{4}(x)</math>, and <math>f_{6}(x)</math>, we may rewrite the expression as | Substituting the definitions of <math>f_{2}(x)</math>, <math>f_{4}(x)</math>, and <math>f_{6}(x)</math>, we may rewrite the expression as | ||
− | <cmath>3(\text{sin}^4 x + \text{cos}^4 x) - 2(\text{sin}^6 x + \text{cos}^6 x) = 1</cmath> | + | <cmath>3(\text{sin}^4{x} + \text{cos}^4{x}) - 2(\text{sin}^6{x} + \text{cos}^6{x}) = 1</cmath> |
We now simplify each term separately using some algebraic manipulation and the Pythagorean identity. | We now simplify each term separately using some algebraic manipulation and the Pythagorean identity. | ||
+ | |||
We can rewrite <math>3(\text{sin}^4 x + \text{cos}^4 x)</math> as | We can rewrite <math>3(\text{sin}^4 x + \text{cos}^4 x)</math> as | ||
− | <math>3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin}^2 x \text{cos}^2 x</math> | + | <math>3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin}^2 x \text{cos}^2 x</math>, which is equivalent to |
+ | <math>3 - 6\text{sin}^2 x \text{cos}^2 x</math>. | ||
+ | |||
+ | As for <math>2(\text{sin}^6 x + \text{cos}^6 x)</math>, we may factor it as | ||
+ | <math>2(\text{sin}^2 x + \text{cos}^2 x)(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)</math> which can be rewritten as | ||
+ | <math>2(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)</math>, and then as | ||
+ | <math>2(\text{sin}^2 x + \text{cos}^2)^2) - 6\text{sin}^2 x \text{cos}^2 x</math>, which is equivalent to | ||
+ | <math>2 - 6\text{sin}^2 x \text{cos}^2 x</math>. | ||
+ | |||
+ | Putting everything together, we have <math>(3 - 6\text{sin}^2 {x} \text{cos}^2 {x}) - (2 - 6\text{sin}^2 {x} \text{cos}^2 {x}) = 1</math> or <math>1 = 1</math>. Therefore, the given equation <math>3f_{4}(x)-2f_{6}(x)=f_{2}(x)</math> is true for all real <math>x</math>, meaning that there are more than <math>8</math> values of <math>x</math> that satisfy the given equation and so the answer is <math>\boxed{\text{(E) more than }8}</math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=P|num-b=9|num-a=11}} | {{AMC12 box|year=2002|ab=P|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:16, 12 July 2024
Problem
Let For how many in is it true that
Solution 1
Divide by 2 on both sides to get Substituting the definitions of , , and , we may rewrite the expression as We now simplify each term separately using some algebraic manipulation and the Pythagorean identity.
We can rewrite as , which is equivalent to .
As for , we may factor it as which can be rewritten as , and then as , which is equivalent to .
Putting everything together, we have or . Therefore, the given equation is true for all real , meaning that there are more than values of that satisfy the given equation and so the answer is .
See also
2002 AMC 12P (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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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