What is the solution of the following trigonometric equation?
\[ \tan(\theta)(\csc(\theta) + 2) = 0 \]
Answer (2)
D : θ = 2 kπ ; k ∈ Z t an θ ( csc θ + 2 ) = 0 ⟺ t an θ = 0 ∨ csc θ + 2 = 0 t an θ = 0 ⟺ θ = kπ ∈ / D csc θ + 2 = 0 csc θ = − 2 s in θ 1 = − 2 − 2 s in θ = 1 / : ( − 2 ) s in θ = − 2 1
( θ = − 6 π + 2 kπ ∨ θ = 6 7 π + 2 kπ ) − an s w er
The solutions to the equation tan ( θ ) ( csc ( θ ) + 2 ) = 0 include angles from tan ( θ ) = 0 which are θ = kπ and from csc ( θ ) = − 2 which are θ = 6 7 π + 2 kπ and θ = 6 11 π + 2 kπ .
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