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

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The equation $\sin^4 x + \cos^7 x = 1$ can be transformed to $\cos^2 x \cdot \left[\cos^5 x - 1 - \sin^2 x\right] = 0$. This leads to two cases: either $\cos^2 x = 0$, giving solutions at $x = \pm \frac{\pi}{2}$, or solving $t^5 + t^2 - 2 = 0$ where $t = \cos x$. The analysis shows that $t^5 + t^2 - 2 \le 0$ for $t \le 1$, indicating that the equation has limited solutions. Further exploration of the transformed equation is needed to determine the total number of solutions within the interval $[-\pi, \pi]$.
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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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