Description
Explanation & Answer
# 3. cos2θ – cosθ = 5 → 2cos2θ – cosθ – 1 = 5 → 2cos2θ – cosθ – 6 = 0
Factor the equation (2cosθ + 3)(cosθ – 2) = 0
Then either 2cosθ + 3 = 0; cosθ = –3/2; this equation does not have a solution.
or cosθ – 2 = 0; cosθ = 2, this equation does not have a solution, too.
Answer: No solutions.
# 4. cos2θ = cosθ → 2cos2θ – cosθ – 1 = 0 → 2cos2θ – cosθ – 1 = 0
Factor the equation (2cosθ + 1)(cosθ – 1) = 0
Then either 2cosθ + 1 = 0; cosθ = –1/2; θ = +– cos–1(–1/2) + 2πk;
θ = +– 2π/3 + 2πk, where k is any integer (k = 0, 1, 2, 3, …, –1, –2, –3, …)
or cosθ – 1 = 0; cosθ = 1, and θ = 2πk , where k is any integer (k = 0, 1, 2, 3, …, –1, –2, –3, …).
Answer: θ = +– 2π/3 + 2πk, where k is any integer (k = 0, 1, 2, 3, …, –1, –2, –3, …)
and θ = 2πk , where k is any integer (k = 0, 1, 2, 3, …, –1, –2, –3, …).