What is the general solution for tan(2t)+tan(t)=1-tan(2t)tan(t) ?

The general solution for tan(2t)+tan(t)=1-tan(2t)tan(t) is t=1.30899…+pin,t=0.26179…+pin

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tan(2t)+tan(t)=1-tan(2t)tan(t)

tan (2 t) + tan (t) = 1 - tan (2 t) tan (t)

t = 1.30899 \dots + πn, t = 0.26179 \dots + πn

Solution steps

tan (2 t) + tan (t) = 1 - tan (2 t) tan (t)

Subtract

1 - tan (2 t) tan (t)

from both sides

tan (2 t) + tan (t) - 1 + tan (2 t) tan (t) = 0

Rewrite using trig identities

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

Solve by substitution

tan (t) = 2 + 3, tan (t) = 2 - 3

tan (t) = 2 + 3 : t = arctan (2 + 3) + πn

tan (t) = 2 - 3 : t = arctan (2 - 3) + πn

Combine all the solutions

t = arctan (2 + 3) + πn, t = arctan (2 - 3) + πn

Show solutions in decimal form

t = 1.30899 \dots + πn, t = 0.26179 \dots + πn

2 cos^{3} (x) - cos (x) = 0

- 2 cos (2 x) = 3

(tan (x) - 1) (2 sin (x) - 3) = 0

tan (θ) (1 + tan^{2} (θ)) = 0

tan (x) = 9

What is the general solution for tan(2t)+tan(t)=1-tan(2t)tan(t) ?
The general solution for tan(2t)+tan(t)=1-tan(2t)tan(t) is t=1.30899…+pin,t=0.26179…+pin