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Difference between revisions of "2002 AMC 12P Problems/Problem 10"

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Divide by 2 on both sides to get <cmath>3f_{4}(x)-2f_{6}(x)=f_{2}(x)</cmath>
 
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
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<cmath>3(\text{sin}^4 x + \text{cos}^4 x) - 2(\text{sin}^6 x + \text{cos}^6 x) = 1</cmath>
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<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.
  
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<math>2 - 6\text{sin}^2 x \text{cos}^2 x</math>.
 
<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 8 values of <math>x</math> that satisfy the given equation and so the answer is <math>\boxed{\text{(E) more than }8}</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 $f_n (x) = \text{sin}^n x + \text{cos}^n x.$ For how many $x$ in $[0,\pi]$ is it true that

\[6f_{4}(x)-4f_{6}(x)=2f_{2}(x)?\]

$\text{(A) }2 \qquad \text{(B) }4 \qquad \text{(C) }6 \qquad \text{(D) }8 \qquad \text{(E) more than }8$

Solution 1

Divide by 2 on both sides to get \[3f_{4}(x)-2f_{6}(x)=f_{2}(x)\] Substituting the definitions of $f_{2}(x)$, $f_{4}(x)$, and $f_{6}(x)$, we may rewrite the expression as \[3(\text{sin}^4{x} + \text{cos}^4{x}) - 2(\text{sin}^6{x} + \text{cos}^6{x}) = 1\] We now simplify each term separately using some algebraic manipulation and the Pythagorean identity.

We can rewrite $3(\text{sin}^4 x + \text{cos}^4 x)$ as $3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin}^2 x \text{cos}^2 x$, which is equivalent to $3 - 6\text{sin}^2 x \text{cos}^2 x$.

As for $2(\text{sin}^6 x + \text{cos}^6 x)$, we may factor it as $2(\text{sin}^2 x + \text{cos}^2 x)(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)$ which can be rewritten as $2(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)$, and then as $2(\text{sin}^2 x + \text{cos}^2)^2) - 6\text{sin}^2 x \text{cos}^2 x$, which is equivalent to $2 - 6\text{sin}^2 x \text{cos}^2 x$.

Putting everything together, we have $(3 - 6\text{sin}^2 {x} \text{cos}^2 {x}) - (2 - 6\text{sin}^2 {x} \text{cos}^2 {x}) = 1$ or $1 = 1$. Therefore, the given equation $3f_{4}(x)-2f_{6}(x)=f_{2}(x)$ is true for all real $x$, meaning that there are more than $8$ values of $x$ that satisfy the given equation and so the answer is $\boxed{\text{(E) more than }8}$.

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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