Solve for x: arctan(x) = arccos(5/13) Solve for [0, 2π)... How?

Feb 26, 2017

$\arctan x = \arccos \left(\frac{5}{13}\right) = {67.38}^{\circ}$ or ${292.62}^{\circ}$ and $x = \pm \frac{12}{5} = \pm 2.4$

Explanation:

$\arctan x$ is the angle whose tangent ratio is $x$.

$\arccos \left(\frac{5}{13}\right)$ is the angle whose cosine ratio is $\frac{5}{13}$.

As we have $\arctan x = \arccos \left(\frac{5}{13}\right)$,

this means the two angles whose tangent ratio is $x$ and the other whose cosine ratio is $\frac{5}{13}$ are identical.

This means $x$ is the tangent of the angle, whose cosine ratio is $\frac{5}{13}$. Let the angle be $A$.

As cosine ratio is $\frac{5}{13}$, we have $\cos A = \frac{5}{13} =$ and hence using a scientific calculator, we get $A = {67.38}^{\circ}$ or ${360}^{\circ} - {67.38}^{\circ} = {292.62}^{\circ}$

i.e. $\arctan x = \arccos \left(\frac{5}{13}\right) = {67.38}^{\circ}$ or ${292.62}^{\circ}$

$\sec A = \frac{13}{5}$ and $\tan A = \sqrt{{\sec}^{2} A - 1} = \sqrt{{\left(\frac{13}{5}\right)}^{2} - 1}$

= $\pm \frac{12}{5}$ i.e. $x = \pm \frac{12}{5}$

Note that while $\tan {67.38}^{\circ} = \frac{12}{5}$, $\tan {292.62}^{\circ} = \frac{12}{5}$