Question

solutions whenever possible. 6cos2(θ) + cos(θ) - 1 = 0 2 - 2sin2(θ) = cos(θ)Solve the following quadratic trig equations algebraically (without graphing technology) for the domain 0 ≤ x ≤ 360°. Use exact solutions whenever possible. A. 6cos2(θ) + cos(θ) - 1 = 0 B. 2 - 2sin2(θ) = cos(θ)

Answer 1

I may be wrong, but I believe this would be the answer:
6cos^2(θ) + cos(θ) - 1 = (2cos(θ) + 1)(3cos(θ) - 1) = 0 

cos(θ) ∈ {-1/2, 1/3} 

cos(θ) = -1/2 ⇒ θ ∈ {2π/3, 4π/3} 

cos(θ) = 1/3 ⇒ θ ∈ {Arccos(1/3), 360° - Arccos(1/3)} ≈ {70.53°, 289.47°} 


2cos²(θ) - cos(θ) = cos(θ)(2cos(θ) - 1) = 0 ⇒ 

cos(θ) ∈ {0, 1/2} ⇒ x ∈ {60°, 90°, 270°, 300°}
Hope this helped and have a great day!