What is the general solution for sec(5 θ/4)=2,0<θ<2pi ?

Q: What is the general solution for sec(5 θ/4)=2,0<θ<2pi ?

The general solution for sec(5 θ/4)=2,0<θ<2pi is θ=(4pi}{15},θ=(4pi)/3 ,θ=\frac{28pi)/(15)

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Popular >

sec(5 θ/4)=2,0<θ<2pi

Solution

sec (5 \frac{θ}{4}) = 2, 0 < θ < 2 π

Solution

θ = \frac{4 π}{15}, θ = \frac{4 π}{3}, θ = \frac{28 π}{15}

Degrees

θ = 4 8^{\circ}, θ = 24 0^{\circ}, θ = 33 6^{\circ}

Solution steps

sec (5 \cdot \frac{θ}{4}) = 2, 0 < θ < 2 π

General solutions for

sec (5 \frac{θ}{4}) = 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}

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

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

Solve

5 \frac{θ}{4} = \frac{π}{3} + 2 πn : θ = \frac{4 π}{15} + \frac{8 πn}{5}

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

Refine

5 \cdot \frac{θ}{4} : \frac{5 θ}{4}

5 \cdot \frac{θ}{4}

Multiply fractions:

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

= \frac{θ \cdot 5}{4}

\frac{5 θ}{4} = \frac{π}{3} + 2 πn

Multiply both sides by

4

\frac{5 θ}{4} = \frac{π}{3} + 2 πn

Multiply both sides by

4

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

Simplify

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

Simplify

\frac{5 θ}{4} \cdot 4 : 5 θ

\frac{5 θ}{4} \cdot 4

Multiply fractions:

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

= \frac{5 θ \cdot 4}{4}

Cancel the common factor:

4

= 5 θ

Simplify

\frac{π}{3} \cdot 4 : \frac{4 π}{3}

\frac{π}{3} \cdot 4

Multiply fractions:

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

= \frac{π 4}{3}

Simplify

2 πn \cdot 4 : 8 πn

2 πn \cdot 4

Multiply the numbers:

2 \cdot 4 = 8

= 8 πn

5 θ = \frac{4 π}{3} + 8 πn

5 θ = \frac{4 π}{3} + 8 πn

5 θ = \frac{4 π}{3} + 8 πn

Divide both sides by

5

5 θ = \frac{4 π}{3} + 8 πn

Divide both sides by

5

\frac{5 θ}{5} = \frac{\frac{4 π}{3}}{5} + \frac{8 πn}{5}

Simplify

\frac{5 θ}{5} = \frac{\frac{4 π}{3}}{5} + \frac{8 πn}{5}

Simplify

\frac{5 θ}{5} : θ

\frac{5 θ}{5}

Divide the numbers:

\frac{5}{5} = 1

= θ

Simplify

\frac{\frac{4 π}{3}}{5} + \frac{8 πn}{5} : \frac{4 π}{15} + \frac{8 πn}{5}

\frac{\frac{4 π}{3}}{5} + \frac{8 πn}{5}

\frac{\frac{4 π}{3}}{5} = \frac{4 π}{15}

\frac{\frac{4 π}{3}}{5}

Apply the fraction rule:

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

= \frac{4 π}{3 \cdot 5}

Multiply the numbers:

3 \cdot 5 = 15

= \frac{4 π}{15}

= \frac{4 π}{15} + \frac{8 πn}{5}

θ = \frac{4 π}{15} + \frac{8 πn}{5}

θ = \frac{4 π}{15} + \frac{8 πn}{5}

θ = \frac{4 π}{15} + \frac{8 πn}{5}

Solve

5 \frac{θ}{4} = \frac{5 π}{3} + 2 πn : θ = \frac{4 π}{3} + \frac{8 πn}{5}

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

Refine

5 \cdot \frac{θ}{4} : \frac{5 θ}{4}

5 \cdot \frac{θ}{4}

Multiply fractions:

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

= \frac{θ \cdot 5}{4}

\frac{5 θ}{4} = \frac{5 π}{3} + 2 πn

Multiply both sides by

4

\frac{5 θ}{4} = \frac{5 π}{3} + 2 πn

Multiply both sides by

4

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

Simplify

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

Simplify

\frac{5 θ}{4} \cdot 4 : 5 θ

\frac{5 θ}{4} \cdot 4

Multiply fractions:

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

= \frac{5 θ \cdot 4}{4}

Cancel the common factor:

4

= 5 θ

Simplify

\frac{5 π}{3} \cdot 4 : \frac{20 π}{3}

\frac{5 π}{3} \cdot 4

Multiply fractions:

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

= \frac{5 π 4}{3}

Multiply the numbers:

5 \cdot 4 = 20

= \frac{20 π}{3}

Simplify

2 πn \cdot 4 : 8 πn

2 πn \cdot 4

Multiply the numbers:

2 \cdot 4 = 8

= 8 πn

5 θ = \frac{20 π}{3} + 8 πn

5 θ = \frac{20 π}{3} + 8 πn

5 θ = \frac{20 π}{3} + 8 πn

Divide both sides by

5

5 θ = \frac{20 π}{3} + 8 πn

Divide both sides by

5

\frac{5 θ}{5} = \frac{\frac{20 π}{3}}{5} + \frac{8 πn}{5}

Simplify

\frac{5 θ}{5} = \frac{\frac{20 π}{3}}{5} + \frac{8 πn}{5}

Simplify

\frac{5 θ}{5} : θ

\frac{5 θ}{5}

Divide the numbers:

\frac{5}{5} = 1

= θ

Simplify

\frac{\frac{20 π}{3}}{5} + \frac{8 πn}{5} : \frac{4 π}{3} + \frac{8 πn}{5}

\frac{\frac{20 π}{3}}{5} + \frac{8 πn}{5}

\frac{\frac{20 π}{3}}{5} = \frac{4 π}{3}

\frac{\frac{20 π}{3}}{5}

Apply the fraction rule:

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

= \frac{20 π}{3 \cdot 5}

Multiply the numbers:

3 \cdot 5 = 15

= \frac{20 π}{15}

Cancel the common factor:

5

= \frac{4 π}{3}

= \frac{4 π}{3} + \frac{8 πn}{5}

θ = \frac{4 π}{3} + \frac{8 πn}{5}

θ = \frac{4 π}{3} + \frac{8 πn}{5}

θ = \frac{4 π}{3} + \frac{8 πn}{5}

θ = \frac{4 π}{15} + \frac{8 πn}{5}, θ = \frac{4 π}{3} + \frac{8 πn}{5}

Solutions for the range

0 < θ < 2 π

θ = \frac{4 π}{15}, θ = \frac{4 π}{3}, θ = \frac{28 π}{15}

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Frequently Asked Questions (FAQ)

What is the general solution for sec(5 θ/4)=2,0<θ<2pi ?
The general solution for sec(5 θ/4)=2,0<θ<2pi is θ=(4pi}{15},θ=(4pi)/3 ,θ=\frac{28pi)/(15)