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# how to calculate x in steps?

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how to calculate x in steps?

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

Guest Feb 22, 2017

#1
+18764
+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
+18764
+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

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