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

The general solution for sqrt(3)sec(2θ)=2 is θ= pi/(12)+pin,θ=(11pi)/(12)+pin

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sqrt(3)sec(2θ)=2

Solution

3 sec (2 θ) = 2

Solution

θ = \frac{π}{12} + πn, θ = \frac{11 π}{12} + πn

Degrees

θ = 1 5^{\circ} + 18 0^{\circ} n, θ = 16 5^{\circ} + 18 0^{\circ} n

Solution steps

3 sec (2 θ) = 2

Divide both sides by

3

3 sec (2 θ) = 2

Divide both sides by

3

\frac{3 sec ( 2 θ )}{3} = \frac{2}{3}

Simplify

\frac{3 sec ( 2 θ )}{3} = \frac{2}{3}

Simplify

\frac{3 sec ( 2 θ )}{3} : sec (2 θ)

\frac{3 sec ( 2 θ )}{3}

Cancel the common factor:

3

= sec (2 θ)

Simplify

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

\frac{2}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{2 3}{3 3}

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{2 3}{3}

sec (2 θ) = \frac{2 3}{3}

sec (2 θ) = \frac{2 3}{3}

sec (2 θ) = \frac{2 3}{3}

General solutions for

sec (2 θ) = \frac{2 3}{3}

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{π}{6} + 2 πn, 2 θ = \frac{11 π}{6} + 2 πn

2 θ = \frac{π}{6} + 2 πn, 2 θ = \frac{11 π}{6} + 2 πn

Solve

2 θ = \frac{π}{6} + 2 πn : θ = \frac{π}{12} + πn

2 θ = \frac{π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{6} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{π}{6}}{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{π}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

θ = \frac{π}{12} + πn

θ = \frac{π}{12} + πn

θ = \frac{π}{12} + πn

Solve

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

2 θ = \frac{11 π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{11 π}{6} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

Simplify

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

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

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

\frac{\frac{11 π}{6}}{2}

Apply the fraction rule:

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

= \frac{11 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{11 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{π}{12} + πn, θ = \frac{11 π}{12} + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)sec(2θ)=2 ?
The general solution for sqrt(3)sec(2θ)=2 is θ= pi/(12)+pin,θ=(11pi)/(12)+pin