What is the general solution for 3sec(2θ)=-sqrt(18) ?

The general solution for 3sec(2θ)=-sqrt(18) is θ=(3pi)/8+pin,θ=(5pi)/8+pin

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3sec(2θ)=-sqrt(18)

Solution

3 sec (2 θ) = - 18

Solution

θ = \frac{3 π}{8} + πn, θ = \frac{5 π}{8} + πn

Solution steps

3 sec (2 θ) = - 18

18 = 32

18

Prime factorization of

18 : 2 \cdot 3^{2}

18

18

divides by

218 = 9 \cdot 2

= 2 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3 \cdot 3

= 2 \cdot 3^{2}

= 3^{2} \cdot 2

Apply radical rule:

= 2 3^{2}

Apply radical rule:

3^{2} = 3

= 32

3 sec (2 θ) = - 32

Divide both sides by

3

3 sec (2 θ) = - 32

Divide both sides by

3

\frac{3 sec ( 2 θ )}{3} = \frac{- 3 2}{3}

Simplify

\frac{3 sec ( 2 θ )}{3} = \frac{- 3 2}{3}

Simplify

\frac{3 sec ( 2 θ )}{3} : sec (2 θ)

\frac{3 sec ( 2 θ )}{3}

Divide the numbers:

\frac{3}{3} = 1

= sec (2 θ)

Simplify

\frac{- 3 2}{3} : - 2

\frac{- 3 2}{3}

Apply the fraction rule:

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

= - \frac{3 2}{3}

Divide the numbers:

\frac{3}{3} = 1

= - 2

sec (2 θ) = - 2

sec (2 θ) = - 2

sec (2 θ) = - 2

General solutions for

sec (2 θ) = - 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}

2 θ = \frac{3 π}{4} + 2 πn, 2 θ = \frac{5 π}{4} + 2 πn

2 θ = \frac{3 π}{4} + 2 πn, 2 θ = \frac{5 π}{4} + 2 πn

Solve

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

2 θ = \frac{3 π}{4} + 2 πn

Divide both sides by

2

2 θ = \frac{3 π}{4} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

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

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

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{3 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{8} + πn

θ = \frac{3 π}{8} + πn

θ = \frac{3 π}{8} + πn

θ = \frac{3 π}{8} + πn

Solve

2 θ = \frac{5 π}{4} + 2 πn : θ = \frac{5 π}{8} + πn

2 θ = \frac{5 π}{4} + 2 πn

Divide both sides by

2

2 θ = \frac{5 π}{4} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2} : \frac{5 π}{8} + πn

\frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{4}}{2} = \frac{5 π}{8}

\frac{\frac{5 π}{4}}{2}

Apply the fraction rule:

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

= \frac{5 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{5 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{8} + πn

θ = \frac{5 π}{8} + πn

θ = \frac{5 π}{8} + πn

θ = \frac{5 π}{8} + πn

θ = \frac{3 π}{8} + πn, θ = \frac{5 π}{8} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3sec(2θ)=-sqrt(18) ?
The general solution for 3sec(2θ)=-sqrt(18) is θ=(3pi)/8+pin,θ=(5pi)/8+pin