Solve equation below:

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

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

Solve equation below:

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

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

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## Comments

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TopNewest\(\textbf{This Solution Only for Real Solution}\)

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

## 1. \(\frac{-pi}{2} \leq z \leq \frac{pi}{2} \)

## 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

\(a^2+aq+t=0\)

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

:) – Pebrudal Zanu · 3 years, 8 months ago

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Hmmmm... – Pebrudal Zanu · 3 years, 8 months ago

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– Mark Mottian · 3 years, 8 months ago

That's an ugly equation!Log in to reply

– Pebrudal Zanu · 3 years, 8 months ago

Yes, I think like that... very ugly :(Log in to reply