What values for θ (0 ≤ θ ≤ 2π) satisfy the equation?
2 sinθ cos θ + √2 cos θ = 0
A. pi/2, 3pi/4, 3pi/2, 7pi/4
B. pi/2, 3pi/4, 5pi/2, 7pi/4
C. pi/2, 5pi/4, 3pi/2, 7pi/4
D. pi/2, 5pi/4, 3pi/4, 7pi/4
Factorizing the equation, we have:
cos θ (2sin θ + √2) = 0
Either cos θ = 0 or 2sin θ + √2 = 0.
If cos θ = 0, then θ = π/2 or 3π/2.
If 2sin θ + √2 = 0, then sin θ = - √2/2, which occurs at 3π/4 and 5π/4.
Therefore, the values of θ that satisfy the equation are π/2, 3π/4, 5π/4, and 3π/2, which is answer choice B.
To solve the equation 2 sinθ cos θ + √2 cos θ = 0, we can factor out the common term cos θ:
cos θ(2 sinθ + √2) = 0
Now we have two possibilities:
1. cos θ = 0:
For this case, θ can be π/2 or 3π/2.
2. 2 sinθ + √2 = 0:
To solve this, we isolate sinθ:
2 sinθ = -√2
sinθ = -√2/2
The two angles that satisfy this condition are π/4 and 5π/4.
Combining the solutions from both cases, our final answer is:
A. π/2, 3π/4, 3π/2, 7π/4