Number of Solutions of $\sin^4 x+\cos^7 x=1$

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Discussion Overview

The discussion centers on determining the number of solutions to the equation $\sin^4 x + \cos^7 x = 1$ for $x \in \left[-\pi, \pi\right]$. Participants explore mathematical reasoning and potential methods for solving the equation.

Discussion Character

  • Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant asks for a lower bound on the number of solutions to the equation.
  • Another participant provides a transformation of the equation, stating that $\cos^7 x = 1 - \sin^4 x$, leading to the expression $\cos^2 x \cdot \left[\cos^5 x - 1 - \sin^2 x\right] = 0$.
  • From this transformation, it is concluded that either $\cos^2 x = 0$ or $\cos^5 x = 1 + \sin^2 x$, resulting in solutions at $x = \pm \frac{\pi}{2}$.
  • A further participant introduces a substitution $t = \cos x$, reformulating the equation to $t^5 + t^2 - 2 = 0$ and discusses the implications of the bounds on $t$.
  • It is noted that $t^5 + t^2 - 2 \le 0$ under the condition $t \le 1$, and that equality holds when $t^5 = t^2 = 1$.

Areas of Agreement / Disagreement

Participants present various approaches and transformations, but there is no consensus on the total number of solutions or how to proceed from the derived equations.

Contextual Notes

The discussion includes assumptions about the bounds of $t$ and the implications of the transformations used, which may affect the interpretation of the solutions.

juantheron
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Total number of solution of $\sin^4 x+\cos^7 x= 1\;,$ Where $x\in \left[-\pi,\pi\right]$
 
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Can you write a lower bound on the number of solutions?
 
Yes Evgeny Makarov,

$\sin^4x+\cos^7x = 1\Rightarrow \cos^7 x = 1-\sin^4 x= \cos^2 x\cdot (1+\sin^2 x)$

So $\cos^2 x \cdot \left[\cos^5 x-1-\sin ^2 x\right] = 0$

So either $\cos^2 x=0$ or $\cos^5 x = 1+\sin^2 x$

So we get $\displaystyle x=\pm \frac{\pi}{2}$ Now how can i solve after that, Help me

Thanks
 
jacks said:
So either $\cos^2 x=0$ or $\cos^5 x = 1+\sin^2 x$

So we get $\displaystyle x=\pm \frac{\pi}{2}$ Now how can i solve after that
Denote $t=\cos x$; then $\sin^2x=1-t^2$ and the equation becomes $t^5+t^2-2=0$. Note that $t\le 1$, so $t^2\le1$ and $t^5\le1$. Therefore, $t^5+t^2-2\le0$ and $t^5+t^2-2=0$ iff $t^5=t^2=1$. Can you finish?
 

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