Asked by Walter
Solve each equation on the given interval. If answers can be exact, leave the exact. If approximations are required, round solutions to the nearest thousandth.
a) -√12 csc (x- pi/3) +7 = 3' XE [-2 pi, 2 pi]
b) 2sin^2 theta = 1 - sin theta, X E [-3pi/2, 2pi]
a) -√12 csc (x- pi/3) +7 = 3' XE [-2 pi, 2 pi]
b) 2sin^2 theta = 1 - sin theta, X E [-3pi/2, 2pi]
Answers
Answered by
mathhelper
-√12 csc (x- pi/3) +7 = 3
-√12 csc (x- pi/3) = -4
csc (x- pi/3) = -4/2√3 = 2/√3
sin (x- pi/3) = √3/2 , so x-π/3 must be in quad I or II
I know sin π/3 = √3/2
then (x-π/3) = π/3 or x-π/3 = π - π/3
x = π/3 + π/3
x = 2π/3
or x-π/3 = π - π/3
x = 2π/3 + π/3 = π
now the period of sin (x- pi/3) is 2π, so another answer would be
2π/3 - 2π = -4π/3
π - 2π = -π
so for the given domain:
x = -4π/3, -π, 2π/3, and π
-√12 csc (x- pi/3) = -4
csc (x- pi/3) = -4/2√3 = 2/√3
sin (x- pi/3) = √3/2 , so x-π/3 must be in quad I or II
I know sin π/3 = √3/2
then (x-π/3) = π/3 or x-π/3 = π - π/3
x = π/3 + π/3
x = 2π/3
or x-π/3 = π - π/3
x = 2π/3 + π/3 = π
now the period of sin (x- pi/3) is 2π, so another answer would be
2π/3 - 2π = -4π/3
π - 2π = -π
so for the given domain:
x = -4π/3, -π, 2π/3, and π
Answered by
mathhelper
2sin^2 θ = 1 - sin θ
This one is easier ...
2sin^2 θ + sin θ - 1 = 0
(2sinθ - 1)(sinθ + 1) = 0
sinθ = 1/2 or sinθ = -1
I will work in degrees for this one, then switch to radians
θ = 30° or θ = 150°
or
θ = 270°
but we have a domains -360° ≤ θ < 360°
so θ = -330°, -210, -90°, 30°, 150°, 270°
in radians:
θ = -11π/6, -7π/6, -π/2, π/6, 5π/6, 3π/2
This one is easier ...
2sin^2 θ + sin θ - 1 = 0
(2sinθ - 1)(sinθ + 1) = 0
sinθ = 1/2 or sinθ = -1
I will work in degrees for this one, then switch to radians
θ = 30° or θ = 150°
or
θ = 270°
but we have a domains -360° ≤ θ < 360°
so θ = -330°, -210, -90°, 30°, 150°, 270°
in radians:
θ = -11π/6, -7π/6, -π/2, π/6, 5π/6, 3π/2
Answered by
GothTutor
Thank you!!
Answered by
GothTutor
I really couldn't solve this one sorry everyone
Answered by
GothTutor
we appreciate you mathhelper
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