rachelmaddie
Full Member
- Joined
- Aug 30, 2019
- Messages
- 851
I need my work checked please.
Solve the equation sinθ + 1 = cos2θ
First, use the formula
cos 2θ = 1 - 2 sin^2(θ)
Replace in terms of sin:
sin(θ) + 1 = 1 - 2sin^2(θ)
Put everything on one side of the equal sign, set it equal to 0, and solve.
2sin^2(θ) + sin(θ) = 0
We can factor out the sin(theta) since it’s common in both terms:
sin(θ)(2sin(θ) + 1) = 0
Because of the zero product property
sin(θ) = 0 or 2sin(θ) + 1 = 0
Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi
θ = π/2, 3π/2,
The next equation needs to be solved for sin(theta)
2sin(θ) + 1 = 0 —> 2sin(θ) = -1 and —> sin(θ) = -½
Find the values of theta where the sin is -½ in the interval.
2π/3, 4π/3
Solution: θ = π/2, 3π/2, 2π/3, 4π/3
Solve the equation sinθ + 1 = cos2θ
First, use the formula
cos 2θ = 1 - 2 sin^2(θ)
Replace in terms of sin:
sin(θ) + 1 = 1 - 2sin^2(θ)
Put everything on one side of the equal sign, set it equal to 0, and solve.
2sin^2(θ) + sin(θ) = 0
We can factor out the sin(theta) since it’s common in both terms:
sin(θ)(2sin(θ) + 1) = 0
Because of the zero product property
sin(θ) = 0 or 2sin(θ) + 1 = 0
Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi
θ = π/2, 3π/2,
The next equation needs to be solved for sin(theta)
2sin(θ) + 1 = 0 —> 2sin(θ) = -1 and —> sin(θ) = -½
Find the values of theta where the sin is -½ in the interval.
2π/3, 4π/3
Solution: θ = π/2, 3π/2, 2π/3, 4π/3
