^{2}Θ=2tan

^{2}Θ+tanΘ+4

^{ }

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

3sec^{2}Θ=2tan^{2}Θ+tanΘ+4^{
}

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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

Thank you for the answer. Im just not understanding how you are going from 3sec^{2}Θ to 3tan^{2}Θ+3 using that identity. would you be able to expand on the first part, the rest i understand.

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## Comments

The top posting shows this problem as:

"i need to solve a trig equatiion

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

"3sec2Θ=2tan2Θ+tanΘ+4"