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

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

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cos(2θ)=sqrt(2)-cos(2θ)

Solution

cos (2 θ) = 2 - cos (2 θ)

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

cos (2 θ) = 2 - cos (2 θ)

Solve by substitution

cos (2 θ) = 2 - cos (2 θ)

Let:

cos (2 θ) = u

u = 2 - u

u = 2 - u : u = \frac{2}{2}

u = 2 - u

Move

u

to the left side

u = 2 - u

Add

u

to both sides

u + u = 2 - u + u

Simplify

2 u = 2

2 u = 2

Divide both sides by

2

2 u = 2

Divide both sides by

2

\frac{2 u}{2} = \frac{2}{2}

Simplify

u = \frac{2}{2}

u = \frac{2}{2}

Substitute back

u = cos (2 θ)

cos (2 θ) = \frac{2}{2}

cos (2 θ) = \frac{2}{2}

cos (2 θ) = \frac{2}{2} : θ = \frac{π}{8} + πn, θ = \frac{7 π}{8} + πn

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

Combine all the solutions

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

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solvefor x,sec(x)=0 solve for

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sin(x/3)=(sqrt(3))/2

sin (\frac{x}{3}) = \frac{3}{2}

Frequently Asked Questions (FAQ)

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