+0

# how to calculate x in steps?

+5
107
1

how to calculate x in steps?

$$\frac{pi}{4}= cos^-1 (\frac{2x^2}{1+x^4})$$

Guest Feb 22, 2017

#1
+18827
+10

how to calculate x in steps?

$$\frac{\pi}{4}= \cos^-1 \left(\frac{2x^2}{1+x^4} \right)$$

$$\begin{array}{|rcll|} \hline \frac{\pi}{4} &=& \cos^-1 \left(\frac{2x^2}{1+x^4} \right) \quad & | \quad \cos() \\ \cos( \frac{\pi}{4} ) &=& \cos(\cos^-1 \left(\frac{2x^2}{1+x^4} \right)) \\ \cos( \frac{\pi}{4} ) &=& \frac{2x^2}{1+x^4} \quad & | \quad \cos( \frac{\pi}{4} ) = \frac{1} {\sqrt{2}} \\ \frac{1} {\sqrt{2}} &=& \frac{2x^2}{1+x^4} \quad & | \quad \cdot (1+x^4) \\ \frac{1} {\sqrt{2}}\cdot (1+x^4) &=& 2x^2 \quad & | \quad \cdot \sqrt{2} \\ 1+x^4 &=& 2\cdot\sqrt{2}\cdot x^2 \quad & | \quad -2\cdot\sqrt{2}\cdot x^2 \\ x^4 -2\cdot\sqrt{2}\cdot x^2 + 1 &=& 0 \\\\ z^2 -2\cdot\sqrt{2}\cdot z + 1 &=& 0 \quad & | \quad z = x^2 \\ z &=& \frac{2\cdot\sqrt{2} \pm \sqrt{(-2\cdot\sqrt{2})^2-4\cdot 1 } } { 2 } \\ z &=& \frac{2\cdot\sqrt{2} \pm \sqrt{8-4 } } { 2 } \\ z &=& \frac{2\cdot\sqrt{2} \pm \sqrt{4 } } { 2 } \\ z &=& \frac{2\cdot\sqrt{2} \pm 2 } { 2 } \\ z &=& \sqrt{2} \pm 1 \quad & | \quad x = \pm\sqrt{z} \\ x &=& \pm\sqrt{ \sqrt{2} \pm 1} \\\\ x_1 &=& + \sqrt{ \sqrt{2} + 1} = 1.55377397403 \\ x_2 &=& + \sqrt{ \sqrt{2} - 1} = 0.64359425291 \\ x_3 &=& - \sqrt{ \sqrt{2} + 1} = -1.55377397403 \\ x_4 &=& - \sqrt{ \sqrt{2} - 1} = -0.64359425291\\ \hline \end{array}$$

heureka  Feb 22, 2017
Sort:

#1
+18827
+10

how to calculate x in steps?

$$\frac{\pi}{4}= \cos^-1 \left(\frac{2x^2}{1+x^4} \right)$$

$$\begin{array}{|rcll|} \hline \frac{\pi}{4} &=& \cos^-1 \left(\frac{2x^2}{1+x^4} \right) \quad & | \quad \cos() \\ \cos( \frac{\pi}{4} ) &=& \cos(\cos^-1 \left(\frac{2x^2}{1+x^4} \right)) \\ \cos( \frac{\pi}{4} ) &=& \frac{2x^2}{1+x^4} \quad & | \quad \cos( \frac{\pi}{4} ) = \frac{1} {\sqrt{2}} \\ \frac{1} {\sqrt{2}} &=& \frac{2x^2}{1+x^4} \quad & | \quad \cdot (1+x^4) \\ \frac{1} {\sqrt{2}}\cdot (1+x^4) &=& 2x^2 \quad & | \quad \cdot \sqrt{2} \\ 1+x^4 &=& 2\cdot\sqrt{2}\cdot x^2 \quad & | \quad -2\cdot\sqrt{2}\cdot x^2 \\ x^4 -2\cdot\sqrt{2}\cdot x^2 + 1 &=& 0 \\\\ z^2 -2\cdot\sqrt{2}\cdot z + 1 &=& 0 \quad & | \quad z = x^2 \\ z &=& \frac{2\cdot\sqrt{2} \pm \sqrt{(-2\cdot\sqrt{2})^2-4\cdot 1 } } { 2 } \\ z &=& \frac{2\cdot\sqrt{2} \pm \sqrt{8-4 } } { 2 } \\ z &=& \frac{2\cdot\sqrt{2} \pm \sqrt{4 } } { 2 } \\ z &=& \frac{2\cdot\sqrt{2} \pm 2 } { 2 } \\ z &=& \sqrt{2} \pm 1 \quad & | \quad x = \pm\sqrt{z} \\ x &=& \pm\sqrt{ \sqrt{2} \pm 1} \\\\ x_1 &=& + \sqrt{ \sqrt{2} + 1} = 1.55377397403 \\ x_2 &=& + \sqrt{ \sqrt{2} - 1} = 0.64359425291 \\ x_3 &=& - \sqrt{ \sqrt{2} + 1} = -1.55377397403 \\ x_4 &=& - \sqrt{ \sqrt{2} - 1} = -0.64359425291\\ \hline \end{array}$$

heureka  Feb 22, 2017

### 13 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details