Solve For Θ
0 ≤ Θ < 360
Quadratic Formula
cos²Θ - 3cosΘ = 3
let cosΘ = x
x² - 3x = 3
x² - 3x - 3 = 0
Quadratic formula:
for ax² + bx +c = 0
x = [-b +/- sqrt(b² - 4ac)] / (2a)
x = [-(-3) +/- sqrt((-3)² - 4(-3))] / (2)(1)
x = [3 +/- sqrt(9 + 12] / (2)
x = [3 +/- sqrt(21)] / (2)
so now we can plug in x = cosΘ.
cosΘ = [3 +/- sqrt(21)] / (2)
Θ = arccos [ (3+sqrt(21))/2 ] and
Θ = arccos [ (3-sqrt(21))/2 ]
but arccos (cos^(-1)) is only defined from -1< x < 1.
( 3/2 + sqrt(21)/2 ) > 1 so this answer can be thrown out.
plugging this into a calculator
Θ = 142.3 degrees
Try completing the square. It will work on this.
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Answers & Comments
Verified answer
cos²Θ - 3cosΘ = 3
let cosΘ = x
x² - 3x = 3
x² - 3x - 3 = 0
Quadratic formula:
for ax² + bx +c = 0
x = [-b +/- sqrt(b² - 4ac)] / (2a)
x = [-(-3) +/- sqrt((-3)² - 4(-3))] / (2)(1)
x = [3 +/- sqrt(9 + 12] / (2)
x = [3 +/- sqrt(21)] / (2)
so now we can plug in x = cosΘ.
cosΘ = [3 +/- sqrt(21)] / (2)
Θ = arccos [ (3+sqrt(21))/2 ] and
Θ = arccos [ (3-sqrt(21))/2 ]
but arccos (cos^(-1)) is only defined from -1< x < 1.
( 3/2 + sqrt(21)/2 ) > 1 so this answer can be thrown out.
Θ = arccos [ (3-sqrt(21))/2 ]
plugging this into a calculator
Θ = 142.3 degrees
Try completing the square. It will work on this.