Find θ in this equation when the domain is restricted 90º ≤ θ ≤ 180º.
A) 135º only
B) 45º and 135º
C) 120º and 180º
D) 90º and 135º
+130511 Write sin2Θ as 2sinΘcosΘ...so we have
2sinΘcosΘ = √(2)cosΘ divide both sides by √(2) =
√(2)sinΘcosΘ = cosΘ subtract cosΘ from both sides
√(2)sinΘcosΘ - cosΘ = 0 factor
cosΘ (√(2)sinΘ - 1) = 0 set both factors to 0
cosΘ = 0 so...... Θ = 90°
and
√(2)sinΘ - 1 = 0 add 1 to both sides
√(2)sinΘ = 1 divide both sides by √(2)
sinΘ = 1/√(2) so.......Θ = 135°
So "D" is the answer.....here's a graph of the solution. here.......https://www.desmos.com/calculator/0zovelnrq5
+130511 Write sin2Θ as 2sinΘcosΘ...so we have
2sinΘcosΘ = √(2)cosΘ divide both sides by √(2) =
√(2)sinΘcosΘ = cosΘ subtract cosΘ from both sides
√(2)sinΘcosΘ - cosΘ = 0 factor
cosΘ (√(2)sinΘ - 1) = 0 set both factors to 0
cosΘ = 0 so...... Θ = 90°
and
√(2)sinΘ - 1 = 0 add 1 to both sides
√(2)sinΘ = 1 divide both sides by √(2)
sinΘ = 1/√(2) so.......Θ = 135°
So "D" is the answer.....here's a graph of the solution. here.......https://www.desmos.com/calculator/0zovelnrq5
