What is the general solution for 6/(3tan(θ)+1)=5 ?

The general solution for 6/(3tan(θ)+1)=5 is θ=0.06656…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6/(3tan(θ)+1)=5

Solution

\frac{6}{3 tan ( θ ) + 1} = 5

Solution

θ = 0.06656 \dots + πn

Degrees

θ = 3.81407 \dots^{\circ} + 18 0^{\circ} n

Solution steps

\frac{6}{3 tan ( θ ) + 1} = 5

Multiply both sides by

3 tan (θ) + 1

\frac{6}{3 tan ( θ ) + 1} = 5

Multiply both sides by

3 tan (θ) + 1

\frac{6}{3 tan ( θ ) + 1} (3 tan (θ) + 1) = 5 (3 tan (θ) + 1)

Simplify

6 = 5 (3 tan (θ) + 1)

6 = 5 (3 tan (θ) + 1)

Switch sides

5 (3 tan (θ) + 1) = 6

Divide both sides by

5

5 (3 tan (θ) + 1) = 6

Divide both sides by

5

\frac{5 ( 3 tan ( θ ) + 1 )}{5} = \frac{6}{5}

Simplify

3 tan (θ) + 1 = \frac{6}{5}

3 tan (θ) + 1 = \frac{6}{5}

Move

1

to the right side

3 tan (θ) + 1 = \frac{6}{5}

Subtract

1

from both sides

3 tan (θ) + 1 - 1 = \frac{6}{5} - 1

Simplify

3 tan (θ) + 1 - 1 = \frac{6}{5} - 1

Simplify

3 tan (θ) + 1 - 1 : 3 tan (θ)

3 tan (θ) + 1 - 1

Add similar elements:

1 - 1 = 0

= 3 tan (θ)

Simplify

\frac{6}{5} - 1 : \frac{1}{5}

\frac{6}{5} - 1

Convert element to fraction:

1 = \frac{1 \cdot 5}{5}

= - \frac{1 \cdot 5}{5} + \frac{6}{5}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 \cdot 5 + 6}{5}

- 1 \cdot 5 + 6 = 1

- 1 \cdot 5 + 6

Multiply the numbers:

1 \cdot 5 = 5

= - 5 + 6

Add/Subtract the numbers:

- 5 + 6 = 1

= 1

= \frac{1}{5}

3 tan (θ) = \frac{1}{5}

3 tan (θ) = \frac{1}{5}

3 tan (θ) = \frac{1}{5}

Divide both sides by

3

3 tan (θ) = \frac{1}{5}

Divide both sides by

3

\frac{3 tan ( θ )}{3} = \frac{\frac{1}{5}}{3}

Simplify

\frac{3 tan ( θ )}{3} = \frac{\frac{1}{5}}{3}

Simplify

\frac{3 tan ( θ )}{3} : tan (θ)

\frac{3 tan ( θ )}{3}

Divide the numbers:

\frac{3}{3} = 1

= tan (θ)

Simplify

\frac{\frac{1}{5}}{3} : \frac{1}{15}

\frac{\frac{1}{5}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{1}{5 \cdot 3}

Multiply the numbers:

5 \cdot 3 = 15

= \frac{1}{15}

tan (θ) = \frac{1}{15}

tan (θ) = \frac{1}{15}

tan (θ) = \frac{1}{15}

Verify Solutions

Find undefined (singularity) points:

tan (θ) = - \frac{1}{3}

Take the denominator(s) of

\frac{6}{3 tan ( θ ) + 1}

and compare to zero

Solve

3 tan (θ) + 1 = 0 : tan (θ) = - \frac{1}{3}

3 tan (θ) + 1 = 0

Move

1

to the right side

3 tan (θ) + 1 = 0

Subtract

1

from both sides

3 tan (θ) + 1 - 1 = 0 - 1

Simplify

3 tan (θ) = - 1

3 tan (θ) = - 1

Divide both sides by

3

3 tan (θ) = - 1

Divide both sides by

3

\frac{3 tan ( θ )}{3} = \frac{- 1}{3}

Simplify

tan (θ) = - \frac{1}{3}

tan (θ) = - \frac{1}{3}

The following points are undefined

tan (θ) = - \frac{1}{3}

Combine undefined points with solutions:

tan (θ) = \frac{1}{15}

Apply trig inverse properties

tan (θ) = \frac{1}{15}

General solutions for

tan (θ) = \frac{1}{15}

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (\frac{1}{15}) + πn

θ = arctan (\frac{1}{15}) + πn

Show solutions in decimal form

θ = 0.06656 \dots + πn

Popular Examples

cos(x)=2-3sin(x)

cos (x) = 2 - 3 sin (x)

tan(x/2)=1,5

tan (\frac{x}{2}) = 1, 5

2cos^2(x)-1-cos(x)=0,0<= x<= 2pi

2 cos^{2} (x) - 1 - cos (x) = 0, 0 \leq x \leq 2 π

cos(θ)=(-3)/5

cos (θ) = \frac{- 3}{5}

tan^2(x)-tan(x)=2

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

Frequently Asked Questions (FAQ)

What is the general solution for 6/(3tan(θ)+1)=5 ?
The general solution for 6/(3tan(θ)+1)=5 is θ=0.06656…+pin