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

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

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-2tan(θ)=-tan^2(θ)-1

- 2 tan (θ) = - tan^{2} (θ) - 1

θ = \frac{π}{4} + πn

Solution steps

- 2 tan (θ) = - tan^{2} (θ) - 1

Solve by substitution

tan (θ) = 1

tan (θ) = 1 : θ = \frac{π}{4} + πn

Combine all the solutions

θ = \frac{π}{4} + πn

sin^{2} (x) + 6 sin (x) = 5

x, y + y = tan (x)

cos (2 x) = \frac{1}{13}

cos (θ) = 0, D = R (θ)

\frac{sin ( θ )}{20} = \frac{sin ( 15 0 ^{\circ} )}{58.1826}

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