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| | ==Problem== | | ==Problem== |
| − | How many angles <math>\theta</math> with <math>0\le\theta\le2\pi</math> satisfy <math>\log(\sin(3\theta))+\log(\cos(2\theta))=0</math>?
| + | Let <math>x</math> be a real number with <math>\sin x \neq -1</math>. What is the sum of the maximum and minimum possible values of <cmath>\frac{(\sin x + \cos x + 1)^{2}}{\sin x + 1}?</cmath> |
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| − | <math> | + | <math>\textbf{(A)}~2 \qquad \textbf{(B)}~3 \qquad \textbf{(C)}~4 \qquad \textbf{(D)}~6 \qquad \textbf{(E)}~8</math> |
| − | \textbf{(A) }0 \qquad | |
| − | \textbf{(B) }1 \qquad | |
| − | \textbf{(C) }2 \qquad | |
| − | \textbf{(D) }3 \qquad | |
| − | \textbf{(E) }4 \qquad | |
| − | </math> | |
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| − | ==Solution 1== | + | ==Solution== |
| − | | + | The expression in question is equal to <cmath>\frac{\sin^{2}x+\cos^{2}x+2\sin x\cos x+2\cos x+2\sin x+1}{\sin x+1}=\frac{2\sin x\cos x+2\sin x+2\cos x+2}{\sin x+1}=\frac{2(\sin x+1)(\cos x+1)}{\sin x+1}.</cmath> Assuming <math>\sin x\neq -1</math>, it is possible to cancel out <math>\sin x+1</math> leaving <math>2\cos x+2</math>. The minimum and maximum values of <math>\cos x</math> are <math>-1</math> and <math>1</math>, respectively, giving <math>2\cdot(-1)+2+2\cdot 1+2=\boxed{\textbf{(C)}~4}</math>. |
| − | Note that this is equivalent to <math>\sin(3\theta)\cos(2\theta)=1</math>, which is clearly only possible when <math>\sin(3\theta)=\cos(2\theta)=\pm1</math>. (If either one is between <math>1</math> and <math>-1</math>, the other one must be greater than <math>1</math> or less than <math>-1</math> to offset the product, which is impossible for sine and cosine.) They cannot be both <math>-1</math> since we cannot take logarithms of negative numbers, so they are both <math>+1</math>. Then <math>3\theta</math> is <math>\dfrac\pi2</math> more than a multiple of <math>2\pi</math> and <math>2\theta</math> is a multiple of <math>2\pi</math>, so <math>\theta</math> is <math>\dfrac\pi6</math> more than a multiple of <math>\dfrac23\pi</math> and also a multiple of <math>\pi</math>. However, a multiple of <math>\dfrac23\pi</math> will always have a denominator of <math>1</math> or <math>3</math>, and never <math>6</math>; it can thus never add with <math>\dfrac\pi6</math> to form an integral multiple of <math>\pi</math>. Thus, there are <math>\boxed{\textbf{(A) }0}</math> solutions.
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| − | | |
| − | ~Technodoggo
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| − | ==Solution 1.1 (less words)==
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| − | <cmath>\log(\sin(3\theta))+\log(\cos(2\theta))=0</cmath>
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| − | <cmath>\log(\sin(3\theta)\cos(2\theta))=0</cmath>
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| − | <cmath>\sin(3\theta)\cos(2\theta)=1</cmath>
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| − | <cmath>\text{Since } -1\le \sin(x),\cos(x)\le 1 \Rightarrow \sin(3\theta)=\cos(2\theta)= \pm1</cmath>
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| − | BUT note that <math>\log(-1)</math> is not real
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| − | <cmath>\Rightarrow \sin(3\theta)=\cos(2\theta)= 1</cmath>
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| − | <cmath>3\theta=\frac{\pi}{2}+2\pi n; \space 2\theta=2\pi m \space (m,n \in \mathbb{Z})</cmath>
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| − | <cmath>\theta=\frac{\pi}{6}+\frac{2\pi n}{3}; \space \theta=\pi m</cmath>
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| − | <cmath>\Rightarrow \theta\text { has no solution}</cmath>
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| − | Giving us <math> \fbox{(A) 0}</math>.
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| − | | |
| − | ~Minor edits by evanhliu2009
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| − | | |
| − | == Solution 2 ==
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| − | Let <math>f(\theta)=\log(\sin(3\theta))</math> and let <math>g(\theta)=\log(\cos(2\theta))</math>.
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| − | | |
| − | Note that <math>-1\leq\sin(3\theta),\cos(2\theta)\leq1</math>. Because the logarithm of a nonpositive number is not real, the functions <math>f</math> and <math>g</math> only exist when <math>\sin(3\theta)</math> and <math>\cos(2\theta)</math> are positive, respectively. Furthermore, because the logarithm of any positive real number less than <math>1</math> is negative, the only case where the function <math>f+g</math> could equal <math>0</math> is if <math>f(\theta)=g(\theta)=0</math>, which only occurs when their respective sine and cosine expressions are both equal to <math>1</math>.
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| − | | |
| − | Thus, we have these two equations, where <math>m</math> and <math>n</math> are any integers:
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| − | \begin{align*}
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| − | \sin(3\theta)=1 \Rightarrow 3\theta = \frac\pi2+2\pi m &\Rightarrow \theta = \frac\pi6+\frac{2\pi}3 m = \frac{4m+1}6\pi\\
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| − | \cos(2\theta)=1 \Rightarrow 2\theta = 2\pi n &\Rightarrow \theta = \pi n | |
| − | \end{align*}
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| − | | |
| − | Because <math>m</math> is an integer, <math>4m+1</math> is odd, and so <math>\frac{4m+1}6</math> is never an integer. Therefore, by the first equation, <math>\theta</math> can never be an integer multiple of pi. Thus, because the second equation requires that <math>\theta</math> be an integer multiple of pi, these two equations cannot both be satisfied, and so there are no solutions to the given equation.
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| − | | |
| − | Thus, we choose answer choice <math>\boxed{\textbf{(A) }0}</math>.
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| − | | |
| − | == Solution 3 (solution 1.1 with graphing) ==
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| − | <cmath>\log(\sin(3\theta))+\log(\cos(2\theta))=0</cmath>
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| − | <cmath>\log(\sin(3\theta))=-\log(\cos(2\theta))=\log(\sec(2\theta))</cmath>
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| − | <cmath>\therefore\text{let } y=\sin(3\theta)=\sec(2\theta), y > 0</cmath>
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| − | Sketching the graphs <math>y=\sin(3x)</math> and <math>y=\sec(2x)</math> for <math>y>0</math>, we see there are no intersections.
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| − | | |
| − | Thus there are <math>\boxed{\textbf{(A) }0}</math> angles that work.
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| − | | |
| − | ~oofitu
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| − | | |
| − | | |
| − | == Video Solution 1 (🚀 3 min solve 🚀) ==
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| − | | |
| − | https://youtu.be/QIilrdvsHtk
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| − | | |
| − | <i>~Education, the Study of Everything</i>
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| | ==See also== | | ==See also== |
| | {{AMC12 box|year=2024|ab=A|num-b=7|num-a=9}} | | {{AMC12 box|year=2024|ab=A|num-b=7|num-a=9}} |
| | {{MAA Notice}} | | {{MAA Notice}} |
be a real number with 
. What is the sum of the maximum and minimum possible values of ![\[\frac{(\sin x + \cos x + 1)^{2}}{\sin x + 1}?\]](http://latex.artofproblemsolving.com/3/d/a/3da8cabdaffda2c1642b692acf9d158dbb9a1cb5.png)
![\[\frac{\sin^{2}x+\cos^{2}x+2\sin x\cos x+2\cos x+2\sin x+1}{\sin x+1}=\frac{2\sin x\cos x+2\sin x+2\cos x+2}{\sin x+1}=\frac{2(\sin x+1)(\cos x+1)}{\sin x+1}.\]](http://latex.artofproblemsolving.com/0/f/a/0faa0359aafdf5ed1d3d525fe4dbb1b2c3ba5bf9.png)
Assuming 
, it is possible to cancel out 
leaving 
. The minimum and maximum values of 
are 
and 
, respectively, giving 
.
