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