What is the general solution for 3(tan(θ)-2)=2tan(θ)-7 ?

The general solution for 3(tan(θ)-2)=2tan(θ)-7 is θ=(3pi)/4+pin

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3(tan(θ)-2)=2tan(θ)-7

3 (tan (θ) - 2) = 2 tan (θ) - 7

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

3 (tan (θ) - 2) = 2 tan (θ) - 7

Solve by substitution

3 (tan (θ) - 2) = 2 tan (θ) - 7

Let:

tan (θ) = u

3 (u - 2) = 2 u - 7

3 (u - 2) = 2 u - 7 : u = - 1

3 (u - 2) = 2 u - 7

Expand

3 (u - 2) : 3 u - 6

3 (u - 2)

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = u, c = 2

= 3 u - 3 \cdot 2

Multiply the numbers:

3 \cdot 2 = 6

= 3 u - 6

3 u - 6 = 2 u - 7

Move

6

to the right side

3 u - 6 = 2 u - 7

Add

6

to both sides

3 u - 6 + 6 = 2 u - 7 + 6

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

3 tan (θ) + 1 = 0, 0 \leq θ \leq 2 π

tan (x) = tan^{2} (x)

y, sin (x y) = x

tan (x) sec (x) = tan (x)

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

What is the general solution for 3(tan(θ)-2)=2tan(θ)-7 ?
The general solution for 3(tan(θ)-2)=2tan(θ)-7 is θ=(3pi)/4+pin