What is the general solution for tan(θ)+12=2tan(θ)+11 ?

The general solution for tan(θ)+12=2tan(θ)+11 is θ= pi/4+pin

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tan(θ)+12=2tan(θ)+11

tan (θ) + 12 = 2 tan (θ) + 11

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

tan (θ) + 12 = 2 tan (θ) + 11

Solve by substitution

tan (θ) + 12 = 2 tan (θ) + 11

Let:

tan (θ) = u

u + 12 = 2 u + 11

u + 12 = 2 u + 11 : u = 1

u + 12 = 2 u + 11

Move

12

to the right side

u + 12 = 2 u + 11

Subtract

12

from both sides

u + 12 - 12 = 2 u + 11 - 12

Simplify

u = 2 u - 1

u = 2 u - 1

Move

2 u

to the left side

u = 2 u - 1

Subtract

2 u

from both sides

u - 2 u = 2 u - 1 - 2 u

Simplify

- u = - 1

- u = - 1

Divide both sides by

- 1

- u = - 1

Divide both sides by

- 1

\frac{- u}{- 1} = \frac{- 1}{- 1}

Simplify

u = 1

u = 1

Substitute back

u = tan (θ)

tan (θ) = 1

tan (θ) = 1

tan (θ) = 1 : θ = \frac{π}{4} + πn

tan (θ) = 1

General solutions for

tan (θ) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Combine all the solutions

θ = \frac{π}{4} + πn

8 sin (x) = 8 tan (x)

9 cot^{2} (x) - 16 = 0

sin (2 x) csc (x) = 1

tan (θ) = 21

2 cos^{2} (t) + cos (t) - 1 = 0

What is the general solution for tan(θ)+12=2tan(θ)+11 ?
The general solution for tan(θ)+12=2tan(θ)+11 is θ= pi/4+pin