What is the general solution for 4tan(θ)=5-tan(θ) ?

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

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4tan(θ)=5-tan(θ)

4 tan (θ) = 5 - tan (θ)

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

4 tan (θ) = 5 - tan (θ)

Solve by substitution

4 tan (θ) = 5 - tan (θ)

Let:

tan (θ) = u

4 u = 5 - u

4 u = 5 - u : u = 1

4 u = 5 - u

Move

u

to the left side

4 u = 5 - u

Add

u

to both sides

4 u + u = 5 - u + u

Simplify

5 u = 5

5 u = 5

Divide both sides by

5

5 u = 5

Divide both sides by

5

\frac{5 u}{5} = \frac{5}{5}

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

2 sec (x) = 1

sin (2 x) = cos (x), 0 \leq < 2 π

x, y^{1} = sin (x)

tan (θ) = \frac{15}{6}

- cos (2 x) = cos^{2} (x) - sin^{2} (x)

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