cos T(2 sin T + sqrt 3) = 0
well right off pi/2 and 3 pi/2
so what about
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
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
2 answers
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
Answer a
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
Answer a