9. 4cos 0 + 1 = 2cos
A. 2π3 , 4π/3
B. π/6, 5π/6
C. 7π/6 , 11π/6
D π/3 , 6π/3
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.
To solve the equation 4cos(θ) + 1 = 2cos(θ), we can rearrange it to form:
4cos(θ) - 2cos(θ) = -1
Combining like terms, we get:
2cos(θ) = -1
Now, let's divide both sides by 2:
cos(θ) = -1/2
Next, we need to find the values of θ that satisfy this equation. We can do this by referring to the unit circle or by using inverse trigonometric functions.
Since the cosine function is negative in the second and third quadrants, we're interested in the values of θ in those quadrants where cos(θ) = -1/2.
From the unit circle, we know that cos(π/3) = 1/2. Thus, to obtain cos(θ) = -1/2, we can subtract π from π/3. Therefore, the solutions lie at:
θ = π - π/3 = 2π/3
Similarly, in the third quadrant, we can subtract π from 5π/3 to obtain:
θ = 5π/3 - π = 2π/3
So, the solutions for θ are 2π/3 and 4π/3.
Hence, the correct answer is:
A. 2π/3, 4π/3