Verified answer

√3sin(θ) - cos(θ) = 1 , where 0 ≤θ≥2π

√3sin(θ) = cos(θ) + 1

3sin^2(θ) = cos^2(θ) + 2cos(θ) + 1

3(1 - cos^2(θ)) = cos^2(θ) + 2cos(θ) + 1

3 - 3cos^2(θ) = cos^2(θ) + 2cos(θ) + 1

0 = 4cos^2(θ) + 2cos(θ) - 2

4cos^2(θ) + 2cos(θ) - 2 = 0

2(2cos^2(θ) + cos(θ) - 1) = 0

2(cos(θ) + 1)(2cos(θ) - 1) = 0

(cos(θ) + 1) = 0

cos(θ) = -1

θ = π

(2cos(θ) - 1) = 0

cos(θ) = 1/2

θ = π/3, 5π/3

√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

√ 3 sin Ө - cos Ө = k cos ( Ө - α ) for k > 0

√ 3 sin Ө - cos Ө = (k cos α) cos Ө + (k sin α) sin Ө

√ 3 = k sin α

- 1 = k cos α

tan α = - √ 3 ----(2nd quadrant)

α = 2 π / 3

3 + 1 = k ²

k = 2

2 cos (Ө - 2 π / 3) = 1

cos (Ө - 2 π / 3) = 1/2

Ө - 2 π / 3 = π / 3 , 5 π / 3

Ө = π , 7 π / 3

Ө = π for 0 ≤ Ө ≤ 2π