What is the general solution for 2(tan(θ)+3)=5+tan(θ) ?

The general solution for 2(tan(θ)+3)=5+tan(θ) is θ=(3pi)/4+pin

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2(tan(θ)+3)=5+tan(θ)

2 (tan (θ) + 3) = 5 + tan (θ)

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

2 (tan (θ) + 3) = 5 + tan (θ)

Solve by substitution

2 (tan (θ) + 3) = 5 + tan (θ)

Let:

tan (θ) = u

2 (u + 3) = 5 + u

2 (u + 3) = 5 + u : u = - 1

2 (u + 3) = 5 + u

Expand

2 (u + 3) : 2 u + 6

2 (u + 3)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = u, c = 3

= 2 u + 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 2 u + 6

2 u + 6 = 5 + u

Move

6

to the right side

2 u + 6 = 5 + u

Subtract

6

from both sides

2 u + 6 - 6 = 5 + u - 6

Simplify

2 u = u - 1

2 u = u - 1

Move

u

to the left side

2 u = u - 1

Subtract

u

from both sides

2 u - u = u - 1 - 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

sin^{2} (x) = \frac{( 9 m - 8 )}{2}

\frac{5}{2} cos (2 x) - \frac{1}{2} = 3 sin (x)

cos (π + x) = \frac{1}{2}

(tan (x) - 4) (16 sin^{2} (x) - 1) = 0

sec^{2} (x) = 2 sin^{2} (x) + tan^{2} (x)

What is the general solution for 2(tan(θ)+3)=5+tan(θ) ?
The general solution for 2(tan(θ)+3)=5+tan(θ) is θ=(3pi)/4+pin