Solve for $x$: $$\mathrm{csc}^{100}x+\tan^{100}x=1$$
I have tried it so many times but couldn't draw any conclusion. Please help.
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@Lucian see here. – Dheeraj Kumar Dec 27 '14 at 12:25
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7All that matters for the question is that $\csc^2 x\ge 1$ and $\tan^2 x \ge 0$ – Mark Bennet Dec 27 '14 at 12:36
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1there doesn't exist any such $x$! – Praveen Dec 27 '14 at 12:59
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According to the Mark Bennet comment: $$\csc x\ge 1 ,\,\,\,\,\,\forall x\in\mathbb{R}.$$ Therefore for real solutions of this trig equation, we need $\tan x=0$ and $\csc x=1.$ If such a solution exist, then$$\sin x=0=1$$ which is a contradiction. Therefore there are no such real $x.$
Bumblebee
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We get, $$-1\leq \sin(x) \leq 1 \implies 0\leq \sin^2(x) \leq 1 \implies 1\leq \csc^2(x) < \infty $$$$\implies 1\leq \csc^{100}(x) < \infty , \hspace{.3cm}\forall x \in \mathbb{R} $$ and $$-\infty < \tan(y)<\infty \implies 0\leq \tan^2(y)<\infty \implies 0\leq \tan^{100}(y)<\infty , \hspace{0.3cm}\forall y \in \mathbb{R}.$$
So the only possibility for, $\csc^{100}(x)+\tan^{100}(x)=1$
is the existence of an $x \in \mathbb{R}$ such that $\csc^{100}(x)=1$ and $\tan^{100}(x)=0$ ,
$ i.e., \sin(x)=\pm 1$ ($\implies x$ is an odd integer multiple of $\frac {\pi}{2}$) and
$\tan(x)=0$ ($\implies x$ is an even integer multiple of $ {\pi}$),
which is not possible.
Praveen
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"not possible" because using the existence of such an $x \in \mathbb{R}$, we can arrive at the contradiction as odd equals even. – Praveen Dec 27 '14 at 13:42
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Note that the given equation suggests $$\mathrm{cosec} x\le 1$$ whihc suggests the only possible solution as $\mathrm{cosec} x=1\Rightarrow x=\frac{\pi}{2} $ which is impossible because that makes $\tan x =\infty$.
Samrat Mukhopadhyay
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