Find the equation of a line through the coordinate (3,−2) and parallel to the line represented by y=12x−2.
1 answer:
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Answer:
θ = (60° )
Step-by-step explanation:
Using the identity
sin²x + cos²x = 1 ⇒ sin²x = 1 - cos²x
Given
cos²θ - sin²θ = 2 - 5cosθ
cos²θ - (1 - cos²θ) = 2 - 5cosθ
cos²θ - 1 + cos²θ = 2 - 5cosθ
2cos²θ - 1 = 2 - 5cosθ ( subtract 2 - 5cosθ from both sides )
2cos²θ + 5cosθ - 3 = 0 ← in standard form
(cosθ + 3)(2cosθ - 1) = 0 ← in factored form
Equate each factor to zero and solve for θ
cosθ + 3 = 0
cosθ = - 3 ← not possible as - 1 ≤ cosθ ≤ 1
2cosθ - 1 = 0
cosθ = , so
θ = (
) =
( or 60° )