Answer:
θ = 150°, 210°
Step-by-step explanation:
Given:
[tex]\cos \theta=- \cos 30^{\circ}, \quad 0 < \theta < 360^{\circ}[/tex]
We know from trigonometric values of special angles that :
[tex]\cos 30^{\circ}=\dfrac{\sqrt{3}}{2}[/tex]
and that 30° is in Quadrant I.
Therefore:
[tex]-\cos 30^{\circ}=-\dfrac{\sqrt{3}}{2}[/tex]
[tex]\begin{array} {|c|c|}\cline{1-2} \sf Quadrant & \sf Reference \: Angle\\\cline{1-2} \rm I & \theta\\\cline{1-2} \rm II & 180^{\circ}-\theta \\\cline{1-2} \rm III & \theta - 180^{\circ} \\\cline{1-2} \rm IV & 360^{\circ} - \theta \\\cline{1-2} \end{array}[/tex]
Cosine is negative In Quadrants II and III. Therefore, use the above table to find the reference angles of 30° that make the cosine of the angle negative:
[tex]\cos (180-30)^{\circ}=-\dfrac{\sqrt{3}}{2} \implies \cos 150^{\circ}=-\dfrac{\sqrt{3}}{2}[/tex]
and:
[tex]\cos (30-180)^{\circ}=-\dfrac{\sqrt{3}}{2} \implies \cos (-150^{\circ})=-\dfrac{\sqrt{3}}{2}[/tex]
Therefore, θ = 150° ± 360°n, -150° ± 360°n.
So the measure of θ in the given interval is 150° and 210°.
