What is the general solution for 9tan(θ)+13=2tan(θ)+6 ?

The general solution for 9tan(θ)+13=2tan(θ)+6 is θ=(3pi)/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

9tan(θ)+13=2tan(θ)+6

9 tan (θ) + 13 = 2 tan (θ) + 6

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

9 tan (θ) + 13 = 2 tan (θ) + 6

Solve by substitution

9 tan (θ) + 13 = 2 tan (θ) + 6

Let:

tan (θ) = u

9 u + 13 = 2 u + 6

9 u + 13 = 2 u + 6 : u = - 1

9 u + 13 = 2 u + 6

Move

13

to the right side

9 u + 13 = 2 u + 6

Subtract

13

from both sides

9 u + 13 - 13 = 2 u + 6 - 13

Simplify

9 u = 2 u - 7

9 u = 2 u - 7

Move

2 u

to the left side

9 u = 2 u - 7

Subtract

2 u

from both sides

9 u - 2 u = 2 u - 7 - 2 u

Simplify

7 u = - 7

7 u = - 7

Divide both sides by

7

7 u = - 7

Divide both sides by

7

\frac{7 u}{7} = \frac{- 7}{7}

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

2 = 3 sin (∣ 5 x ∣) + 1

sin (θ) = - \frac{1}{3}

sin (θ) = - \frac{1}{4}

sin (θ) cos (θ) = \frac{1}{2}

cos (θ) = \frac{- 3}{2}

What is the general solution for 9tan(θ)+13=2tan(θ)+6 ?
The general solution for 9tan(θ)+13=2tan(θ)+6 is θ=(3pi)/4+pin