What is the general solution for tan(2θ)+tan(2θ)=tan(4θ) ?

The general solution for tan(2θ)+tan(2θ)=tan(4θ) is θ=(pin)/2

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tan(2θ)+tan(2θ)=tan(4θ)

tan (2 θ) + tan (2 θ) = tan (4 θ)

θ = \frac{πn}{2}

Solution steps

tan (2 θ) + tan (2 θ) = tan (4 θ)

Subtract

tan (4 θ)

from both sides

2 tan (2 θ) - tan (4 θ) = 0

Let:

u = 2 θ

2 tan (u) - tan (2 u) = 0

Rewrite using trig identities

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

Solve by substitution

tan (u) = 0

tan (u) = 0 : u = πn

Combine all the solutions

u = πn

Substitute back

u = 2 θ

2 θ = πn : θ = \frac{πn}{2}

θ = \frac{πn}{2}

1 - 7 cos (θ) = 8

tanh (x) = \frac{12}{13}

sin (θ) = \frac{5}{7}

8 sin^{2} (x) + 2 sin (x) - 1 = 0

2 sin^{2} (x) = 5 sin (x) - 3

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