# Algebra 2

What values for theta(0 <= theta <= 2pi) satisfy the equation?
2 sin theta cos theta + sqrt 3 cos theta = 0

a. pi/2, 4pi/3, 3pi/2, 5pi/3
b. pi/2, 3pi/4, 3pi/2, 5pi/3
c. pi/2, 3pi/4, 3pi/2, 5pi/4
d. pi/2, pi/4, 3pi/2, 5pi/3

I have spent hours on this last question for my exam and I cannot figure it out. Please help

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1. cos T(2 sin T + sqrt 3) = 0
well right off pi/2 and 3 pi/2
sin T = -sqrt3/2
well then it is a 30,60, 90 triangle and we are talking about the 60 degree corner
sin is - in quadrants 3 and 4
so
pi + pi/3 = 4 pi/3
and 2 pi - pi/3 = 5 pi/3

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2. 2 sin θ cos θ + √3 cos θ = 0

cos θ ( 2 sin θ + √3 ) = 0

Split into two equations:

cos θ = 0

2 sin θ + √3 = 0

1.

cos θ = 0

In interval ≤ θ ≤ 2π

cos θ = 0

for:

θ = π / 2 (90 °)

and

θ = 3π / 2 (270 °)

2.

2 sin θ + √3 = 0 Subtract √3 from both sides

2 sin θ + √3 - √3 = 0 - √3

2 sin θ = - √3 Divide both sides by 2

sin θ = - √3 / 2

In interval ≤ θ ≤ 2π

sin θ = - √3 / 2

for:

θ = 4π / 3 (240 °)

and

θ = 5π / 3 (300 °)

In interval ≤ θ ≤ 2π

2 sin θ cos θ + √3 cos θ = 0

for:

π / 2 , 4π / 3 , 3π / 2 and 5π / 3

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