Equation , range of P

Dec 2015
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Earth
For which values of $P$, does the equation \(\displaystyle \sin^{10}(x)+\cos^{10}(x)=P\) have no solutions?
 
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Jun 2019
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Plot P over $[-\pi,\pi]$ and see the range. My guess is the answer will be $(-\infty,0) \cup (2^{-4},\infty)$.
 
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mathman

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May 2007
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Because of tenth (even) powers, P can never be negative. The maximum is achieved when |sin(x)| or |cos(x)|=1. In that case P=1. The minimum occurs when the terms are equal or $|\sin(x)|=|\cos(x)|=\sqrt{2}/2$. In that case P$=2^{-4}$. This gives a range for P $[2^{-4},1]$.
 
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greg1313

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\(\displaystyle \frac{d}{dx}\left(\sin^{10}(x)+\cos^{10}(x)\right)=10\sin^9(x)\cos(x)-10\cos^9(x)\sin(x)=0\implies\sin(x)=0\,\vee\cos(x)=0\vee\sin(x)=\cos(x)\)

\(\displaystyle \Rightarrow2^{-4}\le P\le1\)
 
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skipjack

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The question asked for the values of $P$ for which the equation has no solutions.
 
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Dec 2015
975
128
Earth
The question asked for the values of $P$ for which the equation has no solutions.
Almost same question , now just avoid the interval of solutions .
\(\displaystyle P\in (-\infty , 1/16) \cup (1,\infty)\).
 
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