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# what is x if the equation is 2(tan-1(x/160)) = tan-1(x/60)?

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what is x if the equation is 2(tan-1(x/160)) = tan-1(x/60)?

Guest Jul 9, 2015

#1
+18843
+10

what is x if the equation is 2(tan-1(x/160)) = tan-1(x/60) ?

$$\small{\text{ \begin{array}{rclrcl} 2 \cdot \left[ \arctan{ \left( \dfrac{ x } { 160 } \right) } \right] &=& \arctan{ \left( \dfrac{ x } { 60 } \right) } \qquad | \qquad \varphi = \arctan{ \left( \dfrac{ x } { 160 } \right) } \\\\ 2\cdot \varphi &=& \arctan{ \left( \dfrac{ x } { 60 } \right) } \qquad |\qquad \tan{()}\\\\ \tan{ (2\cdot \varphi) } &=& \dfrac{x}{60} \\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot\tan{(\varphi)} } { 1-[\tan{(\varphi)]^2} } \\\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot\tan{( \arctan{ \left( \dfrac{ x } { 160 } \right) } )} } { 1-[\tan{( \arctan{ \left( \dfrac{ x } { 160 } \right) } )]^2} } \\\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot \dfrac{ x } { 160 } } { 1- \left( \dfrac{ x } { 160 } \right)^2 } \\\\\ \dfrac{ 2\cdot \dfrac{ x } { 160 } } { 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{x}{60} \\\\ \dfrac{ 1 } { 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{4}{3} \\\\ 1- \left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{3}{4} \\\\ \left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{1}{4} \qquad | \qquad \pm\sqrt{}\\\\ \dfrac{ x } { 160 } &=& \pm0.5 \\\\ x &=& \pm0.5 \cdot 160 \\\\ \mathbf{x_1} & \mathbf{=} & \mathbf{80} \\\\ \mathbf{x_2} & \mathbf{=} & \mathbf{-80} \\\\ \end{array} }}$$

heureka  Jul 10, 2015
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#1
+18843
+10

what is x if the equation is 2(tan-1(x/160)) = tan-1(x/60) ?

$$\small{\text{ \begin{array}{rclrcl} 2 \cdot \left[ \arctan{ \left( \dfrac{ x } { 160 } \right) } \right] &=& \arctan{ \left( \dfrac{ x } { 60 } \right) } \qquad | \qquad \varphi = \arctan{ \left( \dfrac{ x } { 160 } \right) } \\\\ 2\cdot \varphi &=& \arctan{ \left( \dfrac{ x } { 60 } \right) } \qquad |\qquad \tan{()}\\\\ \tan{ (2\cdot \varphi) } &=& \dfrac{x}{60} \\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot\tan{(\varphi)} } { 1-[\tan{(\varphi)]^2} } \\\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot\tan{( \arctan{ \left( \dfrac{ x } { 160 } \right) } )} } { 1-[\tan{( \arctan{ \left( \dfrac{ x } { 160 } \right) } )]^2} } \\\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot \dfrac{ x } { 160 } } { 1- \left( \dfrac{ x } { 160 } \right)^2 } \\\\\ \dfrac{ 2\cdot \dfrac{ x } { 160 } } { 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{x}{60} \\\\ \dfrac{ 1 } { 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{4}{3} \\\\ 1- \left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{3}{4} \\\\ \left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{1}{4} \qquad | \qquad \pm\sqrt{}\\\\ \dfrac{ x } { 160 } &=& \pm0.5 \\\\ x &=& \pm0.5 \cdot 160 \\\\ \mathbf{x_1} & \mathbf{=} & \mathbf{80} \\\\ \mathbf{x_2} & \mathbf{=} & \mathbf{-80} \\\\ \end{array} }}$$

heureka  Jul 10, 2015
#2
+26412
+5

There is also the trivial solution x= 0

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Alan  Jul 10, 2015
#3
+91510
+5

I really like it.