solve cos3x=-0.5

cos(3x) = -0.5

$\small{\text{$ \begin{array}{rcl} \cos{(3x)} &=& -0.5 \qquad | \qquad \pm \arccos{()}\\ 3x &=& \pm \arccos{(-0.5)} \\ x &=& \pm \frac{ \arccos{(-0.5)} } {3} \\\\ x_1 &=& \frac{1}{3} \cdot \arccos{(-0.5)} \\\\ x_1 &=& \frac{1}{3} \cdot 120^{\circ}\\\\ \mathbf{x_1} & \mathbf{=} & \mathbf{ 40^{\circ} } \\\\ x_2 &=& -\frac{1}{3} \cdot \arccos{(-0.5)} \\\\ x_2 &=& -\frac{1}{3} \cdot 120^{\circ} \\\\ x_2 & = & -40^{\circ} \\\\ x_2 &=& -40^{\circ} + 360^{\circ} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{ 320^{\circ} } \\\\ \end{array} $}}$

$\cos \theta = 0.5$ implies an angle of 60 deg, but the negative sign means that the angle lies in the second or third quadrant, (where cosine is negative).

So, $3x=120 \text{ or } 240 \text{ deg, }$or, more generally 120 $\pm$ 360k or 240 $\pm$ 360k deg, where k is any integer.

Dividing by 3,

$x = 40 \pm 120k, \text{ or }80 \pm 120k.$

So,

$x = \dots -80, 40, 160, 280, \dots$

or

$x = \dots -40, 80, 200, 320, \dots .$