What is the general solution for 3tan(θ)+1=2tan(θ) ?

The general solution for 3tan(θ)+1=2tan(θ) is θ=(3pi)/4+pin

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3tan(θ)+1=2tan(θ)

3 tan (θ) + 1 = 2 tan (θ)

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

3 tan (θ) + 1 = 2 tan (θ)

Solve by substitution

3 tan (θ) + 1 = 2 tan (θ)

Let:

tan (θ) = u

3 u + 1 = 2 u

3 u + 1 = 2 u : u = - 1

3 u + 1 = 2 u

Move

1

to the right side

3 u + 1 = 2 u

Subtract

1

from both sides

3 u + 1 - 1 = 2 u - 1

Simplify

3 u = 2 u - 1

3 u = 2 u - 1

Move

2 u

to the left side

3 u = 2 u - 1

Subtract

2 u

from both sides

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

Simplify

u = - 1

u = - 1

Substitute back

u = tan (θ)

tan (θ) = - 1

tan (θ) = - 1

tan (θ) = - 1 : θ = \frac{3 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{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Combine all the solutions

θ = \frac{3 π}{4} + πn

cos (x) = \frac{18}{25}

sin (x) = - \frac{2}{2}, - π \leq x \leq π

w, s (t) = A e^{- c t} cos (wt + b)

0 = 2 sin (t) + cos (2 t)

8 cos (θ) = - 4

What is the general solution for 3tan(θ)+1=2tan(θ) ?
The general solution for 3tan(θ)+1=2tan(θ) is θ=(3pi)/4+pin