What is the general solution for cot(θ-30)= 1/(sqrt(3)) ?

The general solution for cot(θ-30)= 1/(sqrt(3)) is θ=180n+90

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Popular Trigonometry >

cot(θ-30)= 1/(sqrt(3))

Solution

cot (θ - 3 0^{\circ}) = \frac{1}{3}

Solution

θ = 18 0^{\circ} n + 9 0^{\circ}

Radians

θ = \frac{π}{2} + πn

Solution steps

cot (θ - 3 0^{\circ}) = \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}

cot (θ - 3 0^{\circ}) = \frac{3}{3}

General solutions for

cot (θ - 3 0^{\circ}) = \frac{3}{3}

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

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

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

Solve

θ - 3 0^{\circ} = 6 0^{\circ} + 18 0^{\circ} n : θ = 18 0^{\circ} n + 9 0^{\circ}

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

Move

3 0^{\circ}

to the right side

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

Add

3 0^{\circ}

to both sides

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

Simplify

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

Simplify

θ - 3 0^{\circ} + 3 0^{\circ} : θ

θ - 3 0^{\circ} + 3 0^{\circ}

Add similar elements:

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

= θ

Simplify

6 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ} : 18 0^{\circ} n + 9 0^{\circ}

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

Group like terms

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

Least Common Multiplier of

3, 6 : 6

3, 6

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

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

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

3

6

= 3 \cdot 2

Multiply the numbers:

3 \cdot 2 = 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}

= 6 0^{\circ} + 3 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} 2 + 18 0 ^{\circ}}{6}

Add similar elements:

36 0^{\circ} + 18 0^{\circ} = 54 0^{\circ}

= 9 0^{\circ}

Cancel the common factor:

3

= 18 0^{\circ} n + 9 0^{\circ}

θ = 18 0^{\circ} n + 9 0^{\circ}

θ = 18 0^{\circ} n + 9 0^{\circ}

θ = 18 0^{\circ} n + 9 0^{\circ}

θ = 18 0^{\circ} n + 9 0^{\circ}

Graph

Popular Examples

(cot(x)-sqrt(3))(csc(x)+2)=0

sin(x)= 7/(9.4)

cos(x)= 6/11

3cos^2(x)+5sin(x)-1=0

2cos^2(θ)=7cos(θ)-3

Frequently Asked Questions (FAQ)

What is the general solution for cot(θ-30)= 1/(sqrt(3)) ?
The general solution for cot(θ-30)= 1/(sqrt(3)) is θ=180n+90