Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.) 2 sin2(θ) = 2 + cos(2θ)
θ=

Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.)
cos(2θ) + 7 cos(θ) = 8
θ=

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.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.