i need to solve a trig equatiion

Question

solve the equation on the interval 0≤Θ<2π . round answers to 2 decimal places

 

3sec²Θ=2tan²Θ+tanΘ+4

Answer 1

Solve 3sec^2(θ) = 2tan^2(θ) + tan(θ) + 4
on the interval 0≤θ<2π.
Round answers to 2 decimal places.

Pythagorean Identity:

sin^2(θ) + cos^2(θ) = 1

sin^2(θ)/cos^2(θ) + cos^2(θ)/cos^2(θ) = 1/cos^2(θ)

tan^2(θ) + 1 = sec^2(θ)

Substitute tan^2(θ) + 1 for sec^2(θ) in original equation:

3(tan^2(θ) + 1) = 2tan^2(θ) + tan(θ) + 4

3tan^2(θ) + 3 = 2tan^2(θ) + tan(θ) + 4

tan^2(θ) – tan(θ) – 1 = 0

tan(θ) = (1 ± √(1–4(1)(–1)))/2 = (1 ± √(5))/2
≈ -0.6180339887499, 1.6180339887499.

θ ≈ -0.553574358897049, 1.017221967897853

tangent has period of pi, so also these two angles:

θ ≈ pi-0.553574358897049, pi+1.017221967897853
≈ 2.58801829469274, 4.15881462148765.

Add 2 pi to the negative angle:

θ ≈ 2 pi - 0.553574358897049 ≈ 5.72961094828254

Round answers to 2 decimal places:

θ ≈ 1.02, 2.59, 4.16, 5.73 radians