What is the general solution for sec((3θ)/2)=-2,0<= θ<2pi ?

Q: What is the general solution for sec((3θ)/2)=-2,0<= θ<2pi ?

The general solution for sec((3θ)/2)=-2,0<= θ<2pi is θ=(4pi)/9 ,θ=(8pi)/9 ,θ=(16pi)/9

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sec((3θ)/2)=-2,0<= θ<2pi

Solution

sec (\frac{3 θ}{2}) = - 2, 0 \leq θ < 2 π

Solution

θ = \frac{4 π}{9}, θ = \frac{8 π}{9}, θ = \frac{16 π}{9}

Degrees

θ = 8 0^{\circ}, θ = 16 0^{\circ}, θ = 32 0^{\circ}

Solution steps

sec (\frac{3 θ}{2}) = - 2, 0 \leq θ < 2 π

General solutions for

sec (\frac{3 θ}{2}) = - 2

sec (x)

periodicity table with

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

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

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

Solve

\frac{3 θ}{2} = \frac{2 π}{3} + 2 πn : θ = \frac{4 π}{9} + \frac{4 πn}{3}

\frac{3 θ}{2} = \frac{2 π}{3} + 2 πn

Multiply both sides by

2

\frac{3 θ}{2} = \frac{2 π}{3} + 2 πn

Multiply both sides by

2

\frac{2 \cdot 3 θ}{2} = 2 \cdot \frac{2 π}{3} + 2 \cdot 2 πn

Simplify

\frac{2 \cdot 3 θ}{2} = 2 \cdot \frac{2 π}{3} + 2 \cdot 2 πn

Simplify

\frac{2 \cdot 3 θ}{2} : 3 θ

\frac{2 \cdot 3 θ}{2}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6 θ}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3 θ

Simplify

2 \cdot \frac{2 π}{3} + 2 \cdot 2 πn : \frac{4 π}{3} + 4 πn

2 \cdot \frac{2 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{2 π}{3} = \frac{4 π}{3}

2 \cdot \frac{2 π}{3}

Multiply fractions:

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

= \frac{2 π 2}{3}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{4 π}{3} + 4 πn

3 θ = \frac{4 π}{3} + 4 πn

3 θ = \frac{4 π}{3} + 4 πn

3 θ = \frac{4 π}{3} + 4 πn

Divide both sides by

3

3 θ = \frac{4 π}{3} + 4 πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{\frac{4 π}{3}}{3} + \frac{4 πn}{3}

Simplify

\frac{3 θ}{3} = \frac{\frac{4 π}{3}}{3} + \frac{4 πn}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{4 π}{3}}{3} + \frac{4 πn}{3} : \frac{4 π}{9} + \frac{4 πn}{3}

\frac{\frac{4 π}{3}}{3} + \frac{4 πn}{3}

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

\frac{\frac{4 π}{3}}{3}

Apply the fraction rule:

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

= \frac{4 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{4 π}{9}

= \frac{4 π}{9} + \frac{4 πn}{3}

θ = \frac{4 π}{9} + \frac{4 πn}{3}

θ = \frac{4 π}{9} + \frac{4 πn}{3}

θ = \frac{4 π}{9} + \frac{4 πn}{3}

Solve

\frac{3 θ}{2} = \frac{4 π}{3} + 2 πn : θ = \frac{8 π}{9} + \frac{4 πn}{3}

\frac{3 θ}{2} = \frac{4 π}{3} + 2 πn

Multiply both sides by

2

\frac{3 θ}{2} = \frac{4 π}{3} + 2 πn

Multiply both sides by

2

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

Simplify

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

Simplify

\frac{2 \cdot 3 θ}{2} : 3 θ

\frac{2 \cdot 3 θ}{2}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6 θ}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3 θ

Simplify

2 \cdot \frac{4 π}{3} + 2 \cdot 2 πn : \frac{8 π}{3} + 4 πn

2 \cdot \frac{4 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{4 π}{3} = \frac{8 π}{3}

2 \cdot \frac{4 π}{3}

Multiply fractions:

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

= \frac{4 π 2}{3}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{8 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{8 π}{3} + 4 πn

3 θ = \frac{8 π}{3} + 4 πn

3 θ = \frac{8 π}{3} + 4 πn

3 θ = \frac{8 π}{3} + 4 πn

Divide both sides by

3

3 θ = \frac{8 π}{3} + 4 πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{\frac{8 π}{3}}{3} + \frac{4 πn}{3}

Simplify

\frac{3 θ}{3} = \frac{\frac{8 π}{3}}{3} + \frac{4 πn}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{8 π}{3}}{3} + \frac{4 πn}{3} : \frac{8 π}{9} + \frac{4 πn}{3}

\frac{\frac{8 π}{3}}{3} + \frac{4 πn}{3}

\frac{\frac{8 π}{3}}{3} = \frac{8 π}{9}

\frac{\frac{8 π}{3}}{3}

Apply the fraction rule:

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

= \frac{8 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{8 π}{9}

= \frac{8 π}{9} + \frac{4 πn}{3}

θ = \frac{8 π}{9} + \frac{4 πn}{3}

θ = \frac{8 π}{9} + \frac{4 πn}{3}

θ = \frac{8 π}{9} + \frac{4 πn}{3}

θ = \frac{4 π}{9} + \frac{4 πn}{3}, θ = \frac{8 π}{9} + \frac{4 πn}{3}

Solutions for the range

0 \leq θ < 2 π

θ = \frac{4 π}{9}, θ = \frac{8 π}{9}, θ = \frac{16 π}{9}

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Frequently Asked Questions (FAQ)

What is the general solution for sec((3θ)/2)=-2,0<= θ<2pi ?
The general solution for sec((3θ)/2)=-2,0<= θ<2pi is θ=(4pi)/9 ,θ=(8pi)/9 ,θ=(16pi)/9