Show all work
2 cos^2 θ + 11 cos θ + 5 = 0
(2 cos θ + 1)(cos θ + 5) = 0
2 cos θ + 1 = 0 or cos θ + 5 = 0
2 cos θ = -1...............cos θ = -5
cos θ = -1/2................No solution
θ = 120°, 240°, ...
2cos²θ + 11cosθ = -5 2cos²θ + 11cosθ + 5 0 (cosθ + 5)(2cosθ + 1) = 0 (cosθ + 5) = 0 cosθ = -5 Discard (absolute value of cosine could not be > 1) (2cosθ + 1) = 0 cosθ = -1/2 θ = 180° ± 30° = 150°, 210°
2cos²θ + 11cosθ + 5 = 0
(2cosθ + 1)(cosθ + 5) = 0
cosθ = -1/2 or cosθ = -5 ... Reject the second solution.
cosθ = -1/2
θ = 90°(1 + 2k) + 30°(-1)^k, where k is any integer.
2cos^2 θ +11cosθ+5=0
(2cos θ+1) (cosθ+5)=0
(2cos θ+1)=0 and (cosθ+5)=0
cos θ=-1/2 and cosθ=-5
θ=120+n360 but cos θ=-5 is outside of cos domain
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
2 cos^2 θ + 11 cos θ + 5 = 0
(2 cos θ + 1)(cos θ + 5) = 0
2 cos θ + 1 = 0 or cos θ + 5 = 0
2 cos θ = -1...............cos θ = -5
cos θ = -1/2................No solution
θ = 120°, 240°, ...
2cos²θ + 11cosθ = -5 2cos²θ + 11cosθ + 5 0 (cosθ + 5)(2cosθ + 1) = 0 (cosθ + 5) = 0 cosθ = -5 Discard (absolute value of cosine could not be > 1) (2cosθ + 1) = 0 cosθ = -1/2 θ = 180° ± 30° = 150°, 210°
2cos²θ + 11cosθ + 5 = 0
(2cosθ + 1)(cosθ + 5) = 0
cosθ = -1/2 or cosθ = -5 ... Reject the second solution.
cosθ = -1/2
θ = 90°(1 + 2k) + 30°(-1)^k, where k is any integer.
2cos^2 θ +11cosθ+5=0
(2cos θ+1) (cosθ+5)=0
(2cos θ+1)=0 and (cosθ+5)=0
cos θ=-1/2 and cosθ=-5
θ=120+n360 but cos θ=-5 is outside of cos domain