What is the general solution for tan(2θ)-tan(θ)=0,0<θ<2pi ?

The general solution for tan(2θ)-tan(θ)=0,0<θ<2pi is θ=pi

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tan(2θ)-tan(θ)=0,0<θ<2pi

tan (2 θ) - tan (θ) = 0, 0 < θ < 2 π

θ = π

Solution steps

tan (2 θ) - tan (θ) = 0, 0 < θ < 2 π

Rewrite using trig identities

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

Solve by substitution

tan (θ) = 0, tan (θ) = i, tan (θ) = - i

tan (θ) = 0, 0 < θ < 2 π : θ = π

tan (θ) = i, 0 < θ < 2 π :

No Solution

tan (θ) = - i, 0 < θ < 2 π :

No Solution

Combine all the solutions

θ = π

- 8 cos (x) = - 6 sec (x) + 8

cos (2 θ) + sin (θ) = 1

sin (x) - 2 cos (x) = 1

4 cos (x) - 1 = 0

sin (2 θ) = - \frac{1}{2}, 0 \leq θ < 2 π

What is the general solution for tan(2θ)-tan(θ)=0,0<θ<2pi ?
The general solution for tan(2θ)-tan(θ)=0,0<θ<2pi is θ=pi