What is the general solution for 3tan^3(θ)+3tan^2(θ)-tan(θ)-1=0 ?

The general solution for 3tan^3(θ)+3tan^2(θ)-tan(θ)-1=0 is θ=(3pi)/4+pin,θ=(5pi)/6+pin,θ= pi/6+pin

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3tan^3(θ)+3tan^2(θ)-tan(θ)-1=0

3 tan^{3} (θ) + 3 tan^{2} (θ) - tan (θ) - 1 = 0

θ = \frac{3 π}{4} + πn, θ = \frac{5 π}{6} + πn, θ = \frac{π}{6} + πn

Solution steps

3 tan^{3} (θ) + 3 tan^{2} (θ) - tan (θ) - 1 = 0

Solve by substitution

tan (θ) = - 1, tan (θ) = - \frac{3}{3}, tan (θ) = \frac{3}{3}

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

tan (θ) = - \frac{3}{3} : θ = \frac{5 π}{6} + πn

tan (θ) = \frac{3}{3} : θ = \frac{π}{6} + πn

Combine all the solutions

θ = \frac{3 π}{4} + πn, θ = \frac{5 π}{6} + πn, θ = \frac{π}{6} + πn

sin (4 x) + 2 sin (2 x) = 0

cos (2 x) = - \frac{1}{2}

2 sec^{2} (x) + 5 sec (x) + 2 = 0

sin^{2} (θ) = 8 (cos (- θ) + 1)

tan (x) = \frac{7.1}{5}

What is the general solution for 3tan^3(θ)+3tan^2(θ)-tan(θ)-1=0 ?
The general solution for 3tan^3(θ)+3tan^2(θ)-tan(θ)-1=0 is θ=(3pi)/4+pin,θ=(5pi)/6+pin,θ= pi/6+pin