Solve tan2x=sqrt3 for 0<x<360?

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Solve tan2x=sqrt3 for 0<x<360?

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Answer 1 · 2018-06-17T12:12:48

$x \in {15^{\circ}, 105^{\circ}, 195^{\circ}, 285^{\circ}}$ $x in {15^circ, 105^circ, 195^circ,285^circ}$

Explanation:

Let $θ = 2 x$ $theta=2x$
So that the given $tan (2 x) = \sqrt{3}$ $tan(2x)=sqrt(3)$
becomes $tan (θ) = \sqrt{3}$ $tan(theta)=sqrt(3)$

This is one of the standard reference angles, namely $30^{\circ}$ $30^circ$

Based on the CAST quadrant layout we know that this reference angle will apply to Quadrants 1 and 3;

that is to the angles $30^{\circ}$ $30^circ$ and $180^{\circ} + 30^{\circ} = 210^{\circ}$ $180^circ+30^circ=210^circ$
for $θ \in [0 : 360^{\circ}]$ $theta in [0:360^circ]$ i.e. for $x \in [0^{\circ} : 180^{\circ}]$ $x in [0^circ:180^circ]$
and
to the angles $360^{\circ} + 30^{\circ} = 390^{\circ}$ $360^circ+30^circ=390^circ$ and $360^{\circ} + 180^{\circ} + 30^{\circ} = 570^{\circ}$ $360^circ+180^circ+30^circ=570^circ$
for $θ \in (360^{\circ} : 720^{\circ}]$ $theta in (360^circ:720^circ]$ i.e. for $x \in (180^{\circ} : 360^{\circ}]$ $x in (180^circ:360^circ]$

Therefore $θ \in {30^{\circ}, 210^{\circ}, 390^{\circ}, 570^{\circ}}$ $theta in {30^circ,210^circ,390^circ,570^circ}$
and since $2 x = θ$ $2x=theta$
$XXXXX x \in {15^{\circ}, 105^{\circ}, 195^{\circ}, 285^{\circ}}$ $color(white)("XXXXX")x in {15^circ,105^circ,195^circ,285^circ}$

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Solve tan2x=sqrt3 for 0<x<360?

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