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# Algebra Problem

Solve equation below:

$\displaystyle a \sin |2z| + \log_5 (x \sqrt[8]{2 - 5x^8}) + a^2 = 0$

$\displaystyle ((y^2 - 1) \cos^2 z - y \sin 2z + 1)(1 + \sqrt{\pi + 2z} + \sqrt{\pi - 2z}) = 0$

Note by Fariz Azmi Pratama
4 years, 1 month ago

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$$\textbf{This Solution Only for Real Solution}$$

Suppose that:

$$\displaystyle \log_{5} (x(\sqrt[8]{2-5x^8}=t$$

$$\sin |2z|=q$$

Obtained

$$a^2+aq+t=0$$

Using formula $$abc$$ obtained:

$$\displaystyle a=\frac{-q+\sqrt{q^2-4t}}{2}$$

Or:

$$\displaystyle a=\frac{-q-\sqrt{q^2-4t}}{2}$$

From:

http://www.wolframalpha.com/input/?i=y%3Dlog_5%28%28x^8+%282-5x^8%29%29^%281%2F8%29%29

$$t_{max}=-\frac{1}{8}$$

Now, check maximum $$q$$

From equation $$2$$ are obtained

# 2. $$(y^2-1)\cos^2 z-y\sin 2z+1=0$$

$$y \cos z=sin z$$

$$\tan z= y$$,

so we obtain solution $$y=\tan z$$, where $$|z| \leq \frac {\pi}{2}$$

Now, for real solution:

from$$a^2+qa+t$$ first equation we obtain:

$$D \geq 0$$

$$q^2-4t \geq 0$$

And we get maksimum $$t=\frac {-1}{8}$$ from wolframalfa

So,

$$q^2-4t$$, always definit positif for $$0<x < \sqrt[8]{\left ({\frac{2}{5}} \right) }$$

# Solution:

$$y=\tan z$$

$$-\frac{\pi}{2} \leq z \leq \frac{\pi}{2}$$

$$0<x < \sqrt[8]{\left ({\frac{2}{5}} \right) }$$

And

root of

$$a^2+aq+t=0$$

# $$\textbf{I think it's not closed problem}$$

:)

- 4 years, 1 month ago

Hmmmm...

- 4 years, 1 month ago

That's an ugly equation!

- 4 years, 1 month ago

Yes, I think like that... very ugly :(

- 4 years, 1 month ago