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3 votes
What values for θ (0 ≤ θ ≤ 2pie ) satisfy the equation? 

2 sin θ cos θ + Sqr3  cos θ = 0

a.) pie/2, 4pie/3, 3pie/2, 5pie/3

b.) pie/2, 3pie/4, 3pie/2, 5pie/3

c.) pie/2, 3pie/4, 3pie/2, 5pie/4

d.) pie/2, pie/4, 3pie/2, 5pie/3

User Camh
by
5.6k points

1 Answer

4 votes
Factor out the cosθ:
cosθ (2sinθ + sqrt3) = 0
Therefore, the only ways this can happen are if either cosθ = 0 or if (2sinθ + sqrt3) = 0
The first case, cosθ = 0 only at θ = pi/2, 3pi/2.
The second case, (2sinθ + sqrt3) = 0 simplifies to:
sinθ = (-sqrt3)/2
θ = 4pi/3, 5pi/3
Therefore the answer is A.
User Predominant
by
5.5k points
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