What is the general solution for cot(θ+100)=0 ?

The general solution for cot(θ+100)=0 is θ=180n-10

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Popular >

cot(θ+100)=0

Solution

cot (θ + 10 0^{\circ}) = 0

Solution

θ = 18 0^{\circ} n - 1 0^{\circ}

Radians

θ = - \frac{π}{18} + πn

Solution steps

cot (θ + 10 0^{\circ}) = 0

General solutions for

cot (θ + 10 0^{\circ}) = 0

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 + 10 0^{\circ} = 9 0^{\circ} + 18 0^{\circ} n

θ + 10 0^{\circ} = 9 0^{\circ} + 18 0^{\circ} n

Solve

θ + 10 0^{\circ} = 9 0^{\circ} + 18 0^{\circ} n : θ = 18 0^{\circ} n - 1 0^{\circ}

θ + 10 0^{\circ} = 9 0^{\circ} + 18 0^{\circ} n

Move

10 0^{\circ}

to the right side

θ + 10 0^{\circ} = 9 0^{\circ} + 18 0^{\circ} n

Subtract

10 0^{\circ}

from both sides

θ + 10 0^{\circ} - 10 0^{\circ} = 9 0^{\circ} + 18 0^{\circ} n - 10 0^{\circ}

Simplify

θ + 10 0^{\circ} - 10 0^{\circ} = 9 0^{\circ} + 18 0^{\circ} n - 10 0^{\circ}

Simplify

θ + 10 0^{\circ} - 10 0^{\circ} : θ

θ + 10 0^{\circ} - 10 0^{\circ}

Add similar elements:

10 0^{\circ} - 10 0^{\circ} = 0

= θ

Simplify

9 0^{\circ} + 18 0^{\circ} n - 10 0^{\circ} : 18 0^{\circ} n - 1 0^{\circ}

9 0^{\circ} + 18 0^{\circ} n - 10 0^{\circ}

Group like terms

= 18 0^{\circ} n + 9 0^{\circ} - 10 0^{\circ}

Least Common Multiplier of

2, 9 : 18

2, 9

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

9 : 3 \cdot 3

9

9

divides by

39 = 3 \cdot 3

= 3 \cdot 3

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

2

9

= 2 \cdot 3 \cdot 3

Multiply the numbers:

2 \cdot 3 \cdot 3 = 18

= 18

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

18

For

9 0^{\circ} :

multiply the denominator and numerator by

9

9 0^{\circ} = \frac{18 0 ^{\circ} 9}{2 \cdot 9} = 9 0^{\circ}

For

10 0^{\circ} :

multiply the denominator and numerator by

2

10 0^{\circ} = \frac{90 0 ^{\circ} 2}{9 \cdot 2} = 10 0^{\circ}

= 9 0^{\circ} - 10 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} 9 - 180 0 ^{\circ}}{18}

Add similar elements:

162 0^{\circ} - 180 0^{\circ} = - 18 0^{\circ}

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

Apply the fraction rule:

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

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

θ = 18 0^{\circ} n - 1 0^{\circ}

θ = 18 0^{\circ} n - 1 0^{\circ}

θ = 18 0^{\circ} n - 1 0^{\circ}

θ = 18 0^{\circ} n - 1 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for cot(θ+100)=0 ?
The general solution for cot(θ+100)=0 is θ=180n-10