What is the general solution for tan(60-θ)= 1/(sqrt(3)) ?

The general solution for tan(60-θ)= 1/(sqrt(3)) is θ=-180n+30

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tan(60-θ)= 1/(sqrt(3))

Solution

tan (6 0^{\circ} - θ) = \frac{1}{3}

Solution

θ = - 18 0^{\circ} n + 3 0^{\circ}

Radians

θ = \frac{π}{6} - πn

Solution steps

tan (6 0^{\circ} - θ) = \frac{1}{3}

Simplify

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

tan (6 0^{\circ} - θ) = \frac{3}{3}

General solutions for

tan (6 0^{\circ} - θ) = \frac{3}{3}

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

6 0^{\circ} - θ = 3 0^{\circ} + 18 0^{\circ} n

6 0^{\circ} - θ = 3 0^{\circ} + 18 0^{\circ} n

Solve

6 0^{\circ} - θ = 3 0^{\circ} + 18 0^{\circ} n : θ = - 18 0^{\circ} n + 3 0^{\circ}

6 0^{\circ} - θ = 3 0^{\circ} + 18 0^{\circ} n

Move

6 0^{\circ}

to the right side

6 0^{\circ} - θ = 3 0^{\circ} + 18 0^{\circ} n

Subtract

6 0^{\circ}

from both sides

6 0^{\circ} - θ - 6 0^{\circ} = 3 0^{\circ} + 18 0^{\circ} n - 6 0^{\circ}

Simplify

6 0^{\circ} - θ - 6 0^{\circ} = 3 0^{\circ} + 18 0^{\circ} n - 6 0^{\circ}

Simplify

6 0^{\circ} - θ - 6 0^{\circ} : - θ

6 0^{\circ} - θ - 6 0^{\circ}

Add similar elements:

6 0^{\circ} - 6 0^{\circ} = 0

= - θ

Simplify

3 0^{\circ} + 18 0^{\circ} n - 6 0^{\circ} : 18 0^{\circ} n - 3 0^{\circ}

3 0^{\circ} + 18 0^{\circ} n - 6 0^{\circ}

Group like terms

= 18 0^{\circ} n + 3 0^{\circ} - 6 0^{\circ}

Least Common Multiplier of

6, 3 : 6

6, 3

Least Common Multiplier (LCM)

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Multiply each factor the greatest number of times it occurs in either

6

3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

6

For

6 0^{\circ} :

multiply the denominator and numerator by

2

6 0^{\circ} = \frac{18 0 ^{\circ} 2}{3 \cdot 2} = 6 0^{\circ}

= 3 0^{\circ} - 6 0^{\circ}

Since the denominators are equal, combine the fractions:

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

= \frac{18 0 ^{\circ} - 18 0 ^{\circ} 2}{6}

Add similar elements:

18 0^{\circ} - 36 0^{\circ} = - 18 0^{\circ}

= \frac{- 18 0 ^{\circ}}{6}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= 18 0^{\circ} n - 3 0^{\circ}

- θ = 18 0^{\circ} n - 3 0^{\circ}

- θ = 18 0^{\circ} n - 3 0^{\circ}

- θ = 18 0^{\circ} n - 3 0^{\circ}

Divide both sides by

- 1

- θ = 18 0^{\circ} n - 3 0^{\circ}

Divide both sides by

- 1

\frac{- θ}{- 1} = \frac{18 0 ^{\circ} n}{- 1} - \frac{3 0 ^{\circ}}{- 1}

Simplify

\frac{- θ}{- 1} = \frac{18 0 ^{\circ} n}{- 1} - \frac{3 0 ^{\circ}}{- 1}

Simplify

\frac{- θ}{- 1} : θ

\frac{- θ}{- 1}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{θ}{1}

Apply rule

\frac{a}{1} = a

= θ

Simplify

\frac{18 0 ^{\circ} n}{- 1} - \frac{3 0 ^{\circ}}{- 1} : - 18 0^{\circ} n + 3 0^{\circ}

\frac{18 0 ^{\circ} n}{- 1} - \frac{3 0 ^{\circ}}{- 1}

\frac{18 0 ^{\circ} n}{- 1} = - 18 0^{\circ} n

\frac{18 0 ^{\circ} n}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{18 0 ^{\circ} n}{1}

Apply rule

\frac{a}{1} = a

= - 18 0^{\circ} n

= - 18 0^{\circ} n - \frac{3 0 ^{\circ}}{- 1}

\frac{3 0 ^{\circ}}{- 1} = - 3 0^{\circ}

\frac{3 0 ^{\circ}}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{3 0 ^{\circ}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{3 0 ^{\circ}}{1} = 3 0^{\circ}

= - 3 0^{\circ}

= - 18 0^{\circ} n - (- 3 0^{\circ})

Apply rule

- (- a) = a

= - 18 0^{\circ} n + 3 0^{\circ}

θ = - 18 0^{\circ} n + 3 0^{\circ}

θ = - 18 0^{\circ} n + 3 0^{\circ}

θ = - 18 0^{\circ} n + 3 0^{\circ}

θ = - 18 0^{\circ} n + 3 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for tan(60-θ)= 1/(sqrt(3)) ?
The general solution for tan(60-θ)= 1/(sqrt(3)) is θ=-180n+30