(a) Collect the terms in sinα:
2−3=9sinα+21sinα⇒−1=219sinα⇒sinα=−192≈−0.105.
Since arcsin(−192)≈−6.04°, the solutions are
α≈−6.04°+360°k∨α≈186.04°+360°k,k∈Z.
(b) 1+1=−3cosθ−3cosθ⇒2=−6cosθ⇒cosθ=−31. Hence
θ≈±109.47°+360°k,k∈Z.
sinα=−192 (α≈−6.04° ∨ 186.04°);cosθ=−31 (θ≈±109.47°)