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Find all values of θ in the interval 0≤θ<360° that satisfy the equation
2sin^2(θ)+sin(θ)=1
Show work:
thank you.. 10 pts (5 stars) best answer
Find all values of θ in the interval 0≤θ<360° that satisfy the equation
2sin^2(θ)+sin(θ)=1
Show work:
thank you.. 10 pts (5 stars) best answer
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Use the quadratic formula or factor to get (2sin(θ) - 1)(sin(θ) + 1)=0
You should obtain sin(θ)= -1 or sin(θ) = 1/2
The sin(θ) is -1 at 270°, The sin(θ) is 1/2 at 30° and 150°. Take arcsin(1/2) you get 30°, the other angle is 150° because it is above the x-axis and gives a positive trig value.
You should obtain sin(θ)= -1 or sin(θ) = 1/2
The sin(θ) is -1 at 270°, The sin(θ) is 1/2 at 30° and 150°. Take arcsin(1/2) you get 30°, the other angle is 150° because it is above the x-axis and gives a positive trig value.
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Here we go:
2sin^2(θ) + sin(θ) = 1
{sin(θ)}(2sin(θ) + 1) = 1
=>sin(θ) = 1 or 2sin(θ) + 1 = 1
(θ) = sin^-1(1) or 2(θ) = sin^-1(0)
(θ) = 90 degrees or 2(θ) = 0 implies that (θ) = 0 degrees
2sin^2(θ) + sin(θ) = 1
{sin(θ)}(2sin(θ) + 1) = 1
=>sin(θ) = 1 or 2sin(θ) + 1 = 1
(θ) = sin^-1(1) or 2(θ) = sin^-1(0)
(θ) = 90 degrees or 2(θ) = 0 implies that (θ) = 0 degrees