What is the general solution for sqrt(2)cos(2θ)=1 ?

The general solution for sqrt(2)cos(2θ)=1 is θ= pi/8+pin,θ=(7pi)/8+pin

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sqrt(2)cos(2θ)=1

Solution

2 cos (2 θ) = 1

Solution

θ = \frac{π}{8} + πn, θ = \frac{7 π}{8} + πn

Degrees

θ = 22. 5^{\circ} + 18 0^{\circ} n, θ = 157. 5^{\circ} + 18 0^{\circ} n

Solution steps

2 cos (2 θ) = 1

Divide both sides by

2

2 cos (2 θ) = 1

Divide both sides by

2

\frac{2 cos ( 2 θ )}{2} = \frac{1}{2}

Simplify

\frac{2 cos ( 2 θ )}{2} = \frac{1}{2}

Simplify

\frac{2 cos ( 2 θ )}{2} : cos (2 θ)

\frac{2 cos ( 2 θ )}{2}

Cancel the common factor:

2

= cos (2 θ)

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

cos (2 θ) = \frac{2}{2}

cos (2 θ) = \frac{2}{2}

cos (2 θ) = \frac{2}{2}

General solutions for

cos (2 θ) = \frac{2}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Solve

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

2 θ = \frac{π}{4} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{4} + 2 πn

Divide both sides by

2

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

Simplify

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

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

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{8} + πn

θ = \frac{π}{8} + πn

θ = \frac{π}{8} + πn

θ = \frac{π}{8} + πn

Solve

2 θ = \frac{7 π}{4} + 2 πn : θ = \frac{7 π}{8} + πn

2 θ = \frac{7 π}{4} + 2 πn

Divide both sides by

2

2 θ = \frac{7 π}{4} + 2 πn

Divide both sides by

2

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

Simplify

\frac{2 θ}{2} = \frac{\frac{7 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{7 π}{4}}{2} + \frac{2 πn}{2} : \frac{7 π}{8} + πn

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

\frac{\frac{7 π}{4}}{2} = \frac{7 π}{8}

\frac{\frac{7 π}{4}}{2}

Apply the fraction rule:

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

= \frac{7 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{7 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{7 π}{8} + πn

θ = \frac{7 π}{8} + πn

θ = \frac{7 π}{8} + πn

θ = \frac{7 π}{8} + πn

θ = \frac{π}{8} + πn, θ = \frac{7 π}{8} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)cos(2θ)=1 ?
The general solution for sqrt(2)cos(2θ)=1 is θ= pi/8+pin,θ=(7pi)/8+pin