What is the general solution for 4tan(θ)=3+tan(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

4tan(θ)=3+tan(θ)

4 tan (θ) = 3 + tan (θ)

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

4 tan (θ) = 3 + tan (θ)

Solve by substitution

4 tan (θ) = 3 + tan (θ)

Let:

tan (θ) = u

4 u = 3 + u

4 u = 3 + u : u = 1

4 u = 3 + u

Move

u

to the left side

4 u = 3 + u

Subtract

u

from both sides

4 u - u = 3 + u - u

Simplify

3 u = 3

3 u = 3

Divide both sides by

3

3 u = 3

Divide both sides by

3

\frac{3 u}{3} = \frac{3}{3}

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

tan (2 x) + sec (2 x) = 10

so l v e f or a, F = (x^{(\frac{2}{3})}) + (0.9 (3.3 - x^{2})^{\frac{1}{2}}) \cdot sin (a p e x)

2 sin^{2} (x) + 7 cos (x) + 2 = 0

4 sin (\frac{x}{2}) = 4 cos (\frac{x}{2})

tan^{2} (θ) + 8 tan (θ) + 9 = 0

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