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

The general solution for 2cos^2(2θ)=1-cos(2θ) is θ= pi/6+pin,θ=(5pi)/6+pin,θ= pi/2+pin

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2cos^2(2θ)=1-cos(2θ)

Solution

2 cos^{2} (2 θ) = 1 - cos (2 θ)

Solution

θ = \frac{π}{6} + πn, θ = \frac{5 π}{6} + πn, θ = \frac{π}{2} + πn

Degrees

θ = 3 0^{\circ} + 18 0^{\circ} n, θ = 15 0^{\circ} + 18 0^{\circ} n, θ = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

2 cos^{2} (2 θ) = 1 - cos (2 θ)

Solve by substitution

2 cos^{2} (2 θ) = 1 - cos (2 θ)

Let:

cos (2 θ) = u

2 u^{2} = 1 - u

2 u^{2} = 1 - u : u = \frac{1}{2}, u = - 1

2 u^{2} = 1 - u

Move

u

to the left side

2 u^{2} = 1 - u

Add

u

to both sides

2 u^{2} + u = 1 - u + u

Simplify

2 u^{2} + u = 1

2 u^{2} + u = 1

Move

1

to the left side

2 u^{2} + u = 1

Subtract

1

from both sides

2 u^{2} + u - 1 = 1 - 1

Simplify

2 u^{2} + u - 1 = 0

2 u^{2} + u - 1 = 0

Solve with the quadratic formula

2 u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

a = 2, b = 1, c = - 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 1) = 3

1^{2} - 4 \cdot 2 (- 1)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 \cdot 2}, u_{2} = \frac{- 1 - 3}{2 \cdot 2}

u = \frac{- 1 + 3}{2 \cdot 2} : \frac{1}{2}

\frac{- 1 + 3}{2 \cdot 2}

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 1 - 3}{2 \cdot 2} : - 1

\frac{- 1 - 3}{2 \cdot 2}

Subtract the numbers:

- 1 - 3 = - 4

= \frac{- 4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - 1

Substitute back

u = cos (2 θ)

cos (2 θ) = \frac{1}{2}, cos (2 θ) = - 1

cos (2 θ) = \frac{1}{2}, cos (2 θ) = - 1

cos (2 θ) = \frac{1}{2} : θ = \frac{π}{6} + πn, θ = \frac{5 π}{6} + πn

cos (2 θ) = \frac{1}{2}

General solutions for

cos (2 θ) = \frac{1}{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}{3} + 2 πn, 2 θ = \frac{5 π}{3} + 2 πn

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

Solve

2 θ = \frac{π}{3} + 2 πn : θ = \frac{π}{6} + πn

2 θ = \frac{π}{3} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{3} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{π}{3}}{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}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

Solve

2 θ = \frac{5 π}{3} + 2 πn : θ = \frac{5 π}{6} + πn

2 θ = \frac{5 π}{3} + 2 πn

Divide both sides by

2

2 θ = \frac{5 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

\frac{2 θ}{2} = \frac{\frac{5 π}{3}}{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{5 π}{3}}{2} + \frac{2 πn}{2} : \frac{5 π}{6} + πn

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{3}}{2} = \frac{5 π}{6}

\frac{\frac{5 π}{3}}{2}

Apply the fraction rule:

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

= \frac{5 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{5 π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

θ = \frac{π}{6} + πn, θ = \frac{5 π}{6} + πn

cos (2 θ) = - 1 : θ = \frac{π}{2} + πn

cos (2 θ) = - 1

General solutions for

cos (2 θ) = - 1

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 = π + 2 πn

2 θ = π + 2 πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

2 θ = π + 2 πn

Divide both sides by

2

2 θ = π + 2 πn

Divide both sides by

2

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

Simplify

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

Combine all the solutions

θ = \frac{π}{6} + πn, θ = \frac{5 π}{6} + πn, θ = \frac{π}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(2θ)=1-cos(2θ) ?
The general solution for 2cos^2(2θ)=1-cos(2θ) is θ= pi/6+pin,θ=(5pi)/6+pin,θ= pi/2+pin