Subtracting 1 and dividing by 4, we get:
cos 0 = (2cos - 1)/4
Using the identity cos 2θ = 2cos^2θ - 1, we can rewrite the right-hand side as:
cos 0 = cos^2(θ/2)
Taking the square root of both sides, we get:
cos(θ/2) = ±√(cos 0)
Since we want all possible values of θ, we need to consider both plus and minus signs. Using the reference angle of π/3 for √(3)/2, we have:
cos(θ/2) = ±√(3)/2
θ/2 = ±π/6 + 2πk or θ/2 = ±5π/6 + 2πk for some integer k. Multiplying both sides by 2 and simplifying, we get:
θ = π/3 + 4πk or θ = 5π/3 + 4πk, or θ = 7π/3 + 4πk or θ = 11π/3 + 4πk for some integer k.
Therefore, the answer is (C) 7π/6, 11π/6.
9. 4cos 0 + 1 = 2cos
A. 2π3 , 4π/3
B. π/6, 5π/6
C. 7π/6 , 11π/6
D π/3 , 6π/3
1 answer