Solve For Θ
0 ≤ Θ < 360
::
3sinΘ + 3 = 2(1 - sin²Θ)
2sin²Θ + 3sinΘ + 1 = 0
sin²Θ + 3/2sinΘ + 1/2 = 0
(sinΘ + 1/2)(sinΘ + 1) = 0
sinΘ = -1/2, -1
Θ = 330, 270
i'll just use x for theta
subtract cos^2 (x) for both sides
3sinx - 2cos^2x + 3 = 0
Trig identity: cos^2x = 1 - sin^2x
3sinx - 2(1 - sin^2x) + 3 = 0
distribute
3sinx - 2 + 2sin^2x + 3 = 0
2sin^2x + 3sinx + 1 = 0
let y be sinx
2y^2 + 3y = 1 = 0
factor
2y^2 + 2y + 1y + 1 = 0
(2y^2 + 2y) + (1y + 1) = 0
2y (y + 1) + 1(y + 1) = 0
(2y + 1) (y + 1) = 0
y = -1/2 or -1
sinx = -1/2
x = 210 or 330
sinx = -1
x = 270
√3sin(θ) - cos(θ) = 1 3sin^2A = 1 + 2cosA + cos^2A 3(1 -- cos^2A) = 1 + 2cosA + cos^2A 2cos^2A + cosA -- 1 = 0 whence cosA = (--1/4) +/-- 3/4 = 1/2, --1 hence A = pi/3, pi, 5pi/3
Dude, it's summertime. Why you gotta be drillin us already with friggin calculus?! Screw Copernicus.
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
::
3sinΘ + 3 = 2(1 - sin²Θ)
2sin²Θ + 3sinΘ + 1 = 0
sin²Θ + 3/2sinΘ + 1/2 = 0
(sinΘ + 1/2)(sinΘ + 1) = 0
sinΘ = -1/2, -1
Θ = 330, 270
i'll just use x for theta
subtract cos^2 (x) for both sides
3sinx - 2cos^2x + 3 = 0
Trig identity: cos^2x = 1 - sin^2x
3sinx - 2(1 - sin^2x) + 3 = 0
distribute
3sinx - 2 + 2sin^2x + 3 = 0
2sin^2x + 3sinx + 1 = 0
let y be sinx
2y^2 + 3y = 1 = 0
factor
2y^2 + 2y + 1y + 1 = 0
(2y^2 + 2y) + (1y + 1) = 0
2y (y + 1) + 1(y + 1) = 0
(2y + 1) (y + 1) = 0
y = -1/2 or -1
sinx = -1/2
x = 210 or 330
sinx = -1
x = 270
√3sin(θ) - cos(θ) = 1 3sin^2A = 1 + 2cosA + cos^2A 3(1 -- cos^2A) = 1 + 2cosA + cos^2A 2cos^2A + cosA -- 1 = 0 whence cosA = (--1/4) +/-- 3/4 = 1/2, --1 hence A = pi/3, pi, 5pi/3
Dude, it's summertime. Why you gotta be drillin us already with friggin calculus?! Screw Copernicus.