Question

HELP The solutions to the equation 2cosθ−sinθ=0 for 0∘≤θ≤360∘ are

Answer 1

Answer:

θ ∈ { 90 ∘ , 210 ∘ , 270 ∘ , 330 ∘ }

step by step explanation:

Factoring the left side of

2 sin ( θ ) cos ( θ ) + cos ( θ ) = 0

gives

cos ( θ ) ⋅ [ 2 sin ( θ ) + 1 ] = 0

Which implies (for  θ ∈ [ 0 , 360 ∘])

either     cos ( θ ) = 0

           →  θ = 90 ∘ or 270 ∘

OR

2 sin ( θ ) + 1 = 0

→  sin  ( θ )=− 1 2 → θ = 210 ∘ or 330 ∘

→  θ = 210∘ or 330∘