Solve linear trigonometric equations in secant and cosecant

Question

Solve the following equation for θ on the interval [0,2π).

3√3sec(θ)+4=−2

Answer 1

3√3 sec(θ) + 4 = −2

subtract 4 from both sides

3√3 sec(θ) = −6

divide both sides by 3√3

sec(θ) = −6 / 3√3 = − 2 / √3

take the reciprocal of both sides

cos(θ) = − √3 / 2

We know that √3 / 2 = cos (π / 6)

Therefore

θ = π ± (π / 6) or

θ₁ = 5π / 6 and θ₂ = 7π / 6

Answer 2

3sqr3(secT) +4 =-2

3sqr3(secT) = -6

secT =-6/3sqr3 =- 2/sqr3

cosT = 1/secT = -sqr3/2

T = cos^-1(-sqr3/2)

T = 5pi/6 or 7pi/6

Answer 3