What is the general solution for tan(θ/2-pi/6)=-1,0<= θ<2pi ?

Q: What is the general solution for tan(θ/2-pi/6)=-1,0<= θ<2pi ?

The general solution for tan(θ/2-pi/6)=-1,0<= θ<2pi is θ=(11pi)/6

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Popular Trigonometry >

tan(θ/2-pi/6)=-1,0<= θ<2pi

Solution

tan (\frac{θ}{2} - \frac{π}{6}) = - 1, 0 \leq θ < 2 π

Solution

θ = \frac{11 π}{6}

Degrees

θ = 33 0^{\circ}

Solution steps

tan (\frac{θ}{2} - \frac{π}{6}) = - 1, 0 \leq θ < 2 π

General solutions for

tan (\frac{θ}{2} - \frac{π}{6}) = - 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}

\frac{θ}{2} - \frac{π}{6} = \frac{3 π}{4} + πn

\frac{θ}{2} - \frac{π}{6} = \frac{3 π}{4} + πn

Solve

\frac{θ}{2} - \frac{π}{6} = \frac{3 π}{4} + πn : θ = 2 πn + \frac{11 π}{6}

\frac{θ}{2} - \frac{π}{6} = \frac{3 π}{4} + πn

Move

\frac{π}{6}

to the right side

\frac{θ}{2} - \frac{π}{6} = \frac{3 π}{4} + πn

Add

\frac{π}{6}

to both sides

\frac{θ}{2} - \frac{π}{6} + \frac{π}{6} = \frac{3 π}{4} + πn + \frac{π}{6}

Simplify

\frac{θ}{2} - \frac{π}{6} + \frac{π}{6} = \frac{3 π}{4} + πn + \frac{π}{6}

Simplify

\frac{θ}{2} - \frac{π}{6} + \frac{π}{6} : \frac{θ}{2}

\frac{θ}{2} - \frac{π}{6} + \frac{π}{6}

Add similar elements:

- \frac{π}{6} + \frac{π}{6} = 0

= \frac{θ}{2}

Simplify

\frac{3 π}{4} + πn + \frac{π}{6} : πn + \frac{11 π}{12}

\frac{3 π}{4} + πn + \frac{π}{6}

Group like terms

= πn + \frac{π}{6} + \frac{3 π}{4}

Least Common Multiplier of

6, 4 : 12

6, 4

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

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

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

6

4

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

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

12

For

\frac{π}{6} :

multiply the denominator and numerator by

2

\frac{π}{6} = \frac{π 2}{6 \cdot 2} = \frac{π 2}{12}

For

\frac{3 π}{4} :

multiply the denominator and numerator by

3

\frac{3 π}{4} = \frac{3 π 3}{4 \cdot 3} = \frac{9 π}{12}

= \frac{π 2}{12} + \frac{9 π}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 + 9 π}{12}

Add similar elements:

2 π + 9 π = 11 π

= πn + \frac{11 π}{12}

\frac{θ}{2} = πn + \frac{11 π}{12}

\frac{θ}{2} = πn + \frac{11 π}{12}

\frac{θ}{2} = πn + \frac{11 π}{12}

Multiply both sides by

2

\frac{θ}{2} = πn + \frac{11 π}{12}

Multiply both sides by

2

\frac{2 θ}{2} = 2 πn + 2 \cdot \frac{11 π}{12}

Simplify

\frac{2 θ}{2} = 2 πn + 2 \cdot \frac{11 π}{12}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 πn + 2 \cdot \frac{11 π}{12} : 2 πn + \frac{11 π}{6}

2 πn + 2 \cdot \frac{11 π}{12}

2 \cdot \frac{11 π}{12} = \frac{11 π}{6}

2 \cdot \frac{11 π}{12}

Multiply fractions:

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

= \frac{11 π 2}{12}

Multiply the numbers:

11 \cdot 2 = 22

= \frac{22 π}{12}

Cancel the common factor:

2

= \frac{11 π}{6}

= 2 πn + \frac{11 π}{6}

θ = 2 πn + \frac{11 π}{6}

θ = 2 πn + \frac{11 π}{6}

θ = 2 πn + \frac{11 π}{6}

θ = 2 πn + \frac{11 π}{6}

Solutions for the range

0 \leq θ < 2 π

θ = \frac{11 π}{6}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan(θ/2-pi/6)=-1,0<= θ<2pi ?
The general solution for tan(θ/2-pi/6)=-1,0<= θ<2pi is θ=(11pi)/6