Solve 3𝑠𝑖𝑛2𝜃 + 2 cos 𝜃 = 2, for values of 𝜃 such that 0
≤ 𝜃 ≤ 360
We can rewrite the given equation as:
3(1-cos^2θ) + 2cosθ = 2
3 - 3cos^2θ + 2cosθ = 2
3cos^2θ - 2cosθ - 1 = 0
Now let x = cosθ. We get the quadratic equation:
3x^2 - 2x - 1 = 0
We can apply the quadratic formula to solve for x:
x = (2 ± √(2^2 - 4(3)(-1)))/(2(3))
x = (2 ± √(4 + 12))/(6)
x = (2 ± √(16))/6
x = (2 ± 4)/6
x = 1, x = -1/3
Now we find the corresponding values of θ for each value of x:
For x = cosθ = 1, the only value of θ in the range [0, 360] is θ = 0 degrees.
For x = cosθ = -1/3, θ = ± arccos(-1/3) ≈ ±109.471°. Since θ is only allowed to be non-negative in this problem, we disregard the negative solution and are left with θ ≈ 109.471°.
So, the solutions for θ in the given range are:
θ = 0°, θ ≈ 109.471°.