Answer:
We can start by using the identity 2 sin ϕ cos ϕ = sin 2ϕ to simplify the left-hand side of the equation:
2 sin ϕ cos ϕ + 2 sin ϕ + cos ϕ + 1 = sin 2ϕ + 2 sin ϕ + cos ϕ + 1
Next, we can use another identity, sin^2 ϕ + cos^2 ϕ = 1, to eliminate the cosine term:
sin 2ϕ + 2 sin ϕ + cos ϕ + 1 = sin 2ϕ + 2 sin ϕ + √(1 - sin^2 ϕ) + 1
Now, we can substitute u = sin ϕ to obtain a quadratic equation in u:
sin 2ϕ + 2 sin ϕ + √(1 - sin^2 ϕ) + 1 = sin^2 ϕ + 2 sin ϕ + √(1 - sin^2 ϕ) + 1
sin 2ϕ = sin^2 ϕ
2 sin ϕ cos ϕ = sin^2 ϕ - cos^2 ϕ
2 u √(1 - u^2) = u^2 - (1 - u^2)
2 u √(1 - u^2) = 2 u^2 - 1
4 u^2 (1 - u^2) = (2 u^2 - 1)^2
4 u^4 - 4 u^2 + 1 = 0
This is a quadratic equation in u^2, which can be solved using the quadratic formula:
u^2 = [4 ± √(16 - 16)] / 8 = 1/2 or u^2 = 1
Since 0 < ϕ < 2π, we have 0 < u < 1, so u^2 = 1/2. Therefore, sin ϕ = u = ±√(1/2) and cos ϕ = ±√(1 - u^2) = ±√(1/2).
We can summarize the solutions as follows:
sin ϕ = √(1/2) or sin ϕ = -√(1/2)
cos ϕ = √(1/2) or cos ϕ = -√(1/2)
Therefore