We have to apply the inverse trigonometric functions to solve this equations:
a)
[tex]\begin{gathered} 6\sin (\theta)=1 \\ \sin (\theta)=\frac{1}{6} \\ \theta=\arcsin (\frac{1}{6}) \\ \theta_1\approx0.167 \\ \theta_2=\pi-0.167\approx2.974 \end{gathered}[/tex]b)
[tex]\begin{gathered} 9\cdot\sin (\theta)=9 \\ \sin (\theta)=1 \\ \theta=\arcsin (1)=\frac{\pi}{2}\approx1.571 \end{gathered}[/tex]c)
[tex]\begin{gathered} 5\cdot\sin (\theta)=13 \\ \sin (\theta)=\frac{13}{5} \\ \theta=\arcsin (\frac{13}{5})\longrightarrow\text{ no solution} \end{gathered}[/tex]Answer:
a) θ = 0.167, 2.974
b) θ = 1.571
c) θ = DNE