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For the first equation, recall that sin²(θ) = (1 - cos(2θ))/2. Then
2 sin²(θ) = 2 + cos(2θ)
1 - cos(2θ) = 2 + cos(2θ)
2 cos(2θ) = -1
cos(2θ) = -1/2
2θ = arccos(-1/2) + 2nπ or 2θ = 2π - arccos(-1/2) + 2nπ
(where n is any integer)
2θ = 2π/3 + 2nπ or 2θ = 4π/3 + 2nπ
θ = π/3 + nπ or θ = 2π/3 + nπ
In the interval [0, 2π), the solutions are θ = π/3, 2π/3, 4π/3, 5π/3.
For the second equation, rearrange the previous identity to arrive at
cos(2θ) = 1 - 2 sin²(θ) = 2 cos²(θ) - 1
Then
cos(2θ) + 7 cos(θ) = 8
2 cos²(θ) - 1 + 7 cos(θ) = 8
2 cos²(θ) + 7 cos(θ) - 9 = 0
(2 cos(θ) + 9) (cos(θ) - 1) = 0
2 cos(θ) + 9 = 0 or cos(θ) - 1 = 0
cos(θ) = -9/2 or cos(θ) = 1
Since |-9/2| > 1, and cos(θ) is bounded between -1 and 1, the first case offers no solutions. This leaves us with
cos(θ) = 1
θ = arccos(1) + 2nπ
θ = 2nπ
so that there is only one solution in [0, 2π), θ = 0.