What is the general solution for -2tan^2(θ)+2tan(θ)+1=3tan(θ) ?

The general solution for -2tan^2(θ)+2tan(θ)+1=3tan(θ) is θ=(3pi)/4+pin,θ=0.46364…+pin

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-2tan^2(θ)+2tan(θ)+1=3tan(θ)

- 2 tan^{2} (θ) + 2 tan (θ) + 1 = 3 tan (θ)

θ = \frac{3 π}{4} + πn, θ = 0.46364 \dots + πn

Solution steps

- 2 tan^{2} (θ) + 2 tan (θ) + 1 = 3 tan (θ)

Solve by substitution

tan (θ) = - 1, tan (θ) = \frac{1}{2}

tan (θ) = - 1 : θ = \frac{3 π}{4} + πn

tan (θ) = \frac{1}{2} : θ = arctan (\frac{1}{2}) + πn

Combine all the solutions

θ = \frac{3 π}{4} + πn, θ = arctan (\frac{1}{2}) + πn

Show solutions in decimal form

θ = \frac{3 π}{4} + πn, θ = 0.46364 \dots + πn

cos (x) = \frac{60}{61}, tan (x) = \frac{11}{60}

sin (x) = - 2 sin (x)

19 cot^{2} (x) - 19 = 0

θ, 2 sin (θ) = csc (θ) + 1

sin^{2} (x) \cdot 2 cos (x) = 2

What is the general solution for -2tan^2(θ)+2tan(θ)+1=3tan(θ) ?
The general solution for -2tan^2(θ)+2tan(θ)+1=3tan(θ) is θ=(3pi)/4+pin,θ=0.46364…+pin