# 3. cos2θ – cosθ = 5 → 2cos^{2}θ – cosθ – 1 = 5 → 2cos^{2}θ – 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 → 2cos^{2}θ – cosθ – 1 = 0 → 2cos^{2}θ – 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, …).

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