What values for θ (0 ≤ θ ≤ 2pi) satisfy the equation?
2 sin θ cos θ + cos θ = 0
We can factor out cos θ from the left-hand side of the equation:
2 sin θ cos θ + cos θ = cos θ (2 sin θ + 1) = 0
This means that either cos θ = 0 or 2 sin θ + 1 = 0.
If cos θ = 0, then θ must be either π/2 or 3π/2.
If 2 sin θ + 1 = 0, then sin θ = -1/2. There are two values of θ between 0 and 2π that satisfy this condition: θ = 7π/6 and θ = 11π/6.
Therefore, the values of θ that satisfy the equation are θ = π/2, 3π/2, 7π/6, and 11π/6.
To solve the equation 2 sin θ cos θ + cos θ = 0, we can factor out cos θ from the left side of the equation.
cos θ(2sin θ + 1) = 0
Now, we can set each factor equal to zero and solve for θ.
cos θ = 0
To determine the values of θ that satisfy this equation, we can look at the unit circle or the graph of the cosine function. On the unit circle, the cosine is equal to zero at θ = π/2 and θ = 3π/2.
2sin θ + 1 = 0
Now, we can solve this equation for θ. Subtracting 1 from both sides:
2sin θ = -1
Next, divide by 2:
sin θ = -1/2
The values of sin θ equal to -1/2 are θ = 7π/6 and θ = 11π/6.
Therefore, the values for θ (0 ≤ θ ≤ 2π) that satisfy the equation 2 sin θ cos θ + cos θ = 0 are θ = π/2, 3π/2, 7π/6, and 11π/6.