Answer:
0° and 18°
Step-by-step explanation:
3tan²β - tanβ = 0
tanβ(3tanβ - 1) = 0
tan β = 0 ⇒ [tex]\beta _{1}[/tex] = 0°
3tan β - 1 = 0 ⇒ tan β = [tex]\frac{1}{3}[/tex] ⇒ [tex]\beta _{2}[/tex] = arctan ( [tex]\frac{1}{3}[/tex] ) ≈ 18°
[tex]3 \tan {}^{2} \theta - \tan\theta= 0 \\ \tan\theta(3 \tan \theta- 1) = 0✓\\\\\tan\theta = 0 \\\boxed{\theta = 0°} ✓\\\\ 3 \tan\theta - 1 = 0 \\3 \tan \theta = 1 \\ \tan \theta = \frac{1}{3} \\ \theta = { \tan }^{ - 1} ( \frac{1}{3} ) \\\boxed{\theta = 18.4°}✓[/tex]
