What is the general solution for sec(4θ)=sqrt(2) ?

The general solution for sec(4θ)=sqrt(2) is θ= pi/(16)+(pin)/2 ,θ=(7pi}{16}+\frac{pin)/2

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Popular >

sec(4θ)=sqrt(2)

Solution

sec (4 θ) = 2

Solution

θ = \frac{π}{16} + \frac{πn}{2}, θ = \frac{7 π}{16} + \frac{πn}{2}

Degrees

θ = 11.2 5^{\circ} + 9 0^{\circ} n, θ = 78.7 5^{\circ} + 9 0^{\circ} n

Solution steps

sec (4 θ) = 2

General solutions for

sec (4 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}

4 θ = \frac{π}{4} + 2 πn, 4 θ = \frac{7 π}{4} + 2 πn

4 θ = \frac{π}{4} + 2 πn, 4 θ = \frac{7 π}{4} + 2 πn

Solve

4 θ = \frac{π}{4} + 2 πn : θ = \frac{π}{16} + \frac{πn}{2}

4 θ = \frac{π}{4} + 2 πn

Divide both sides by

4

4 θ = \frac{π}{4} + 2 πn

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

Divide the numbers:

\frac{4}{4} = 1

= θ

Simplify

\frac{\frac{π}{4}}{4} + \frac{2 πn}{4} : \frac{π}{16} + \frac{πn}{2}

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

\frac{\frac{π}{4}}{4} = \frac{π}{16}

\frac{\frac{π}{4}}{4}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 4}

Multiply the numbers:

4 \cdot 4 = 16

= \frac{π}{16}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{π}{16} + \frac{πn}{2}

θ = \frac{π}{16} + \frac{πn}{2}

θ = \frac{π}{16} + \frac{πn}{2}

θ = \frac{π}{16} + \frac{πn}{2}

Solve

4 θ = \frac{7 π}{4} + 2 πn : θ = \frac{7 π}{16} + \frac{πn}{2}

4 θ = \frac{7 π}{4} + 2 πn

Divide both sides by

4

4 θ = \frac{7 π}{4} + 2 πn

Divide both sides by

4

\frac{4 θ}{4} = \frac{\frac{7 π}{4}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 θ}{4} = \frac{\frac{7 π}{4}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

Divide the numbers:

\frac{4}{4} = 1

= θ

Simplify

\frac{\frac{7 π}{4}}{4} + \frac{2 πn}{4} : \frac{7 π}{16} + \frac{πn}{2}

\frac{\frac{7 π}{4}}{4} + \frac{2 πn}{4}

\frac{\frac{7 π}{4}}{4} = \frac{7 π}{16}

\frac{\frac{7 π}{4}}{4}

Apply the fraction rule:

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

= \frac{7 π}{4 \cdot 4}

Multiply the numbers:

4 \cdot 4 = 16

= \frac{7 π}{16}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{7 π}{16} + \frac{πn}{2}

θ = \frac{7 π}{16} + \frac{πn}{2}

θ = \frac{7 π}{16} + \frac{πn}{2}

θ = \frac{7 π}{16} + \frac{πn}{2}

θ = \frac{π}{16} + \frac{πn}{2}, θ = \frac{7 π}{16} + \frac{πn}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for sec(4θ)=sqrt(2) ?
The general solution for sec(4θ)=sqrt(2) is θ= pi/(16)+(pin)/2 ,θ=(7pi}{16}+\frac{pin)/2