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

The general solution for -1=2cos(2θ) is θ= pi/3+pin,θ=(2pi)/3+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-1=2cos(2θ)

Solution

- 1 = 2 cos (2 θ)

Solution

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

Degrees

θ = 6 0^{\circ} + 18 0^{\circ} n, θ = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

- 1 = 2 cos (2 θ)

Switch sides

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

cos (2 θ) = - \frac{1}{2}

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

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

Solve

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

2 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

2

2 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

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

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 π}{6}

Cancel the common factor:

2

= \frac{π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{3} + πn

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

Solve

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

2 θ = \frac{4 π}{3} + 2 πn

Divide both sides by

2

2 θ = \frac{4 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

\frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{4 π}{3}}{2} = \frac{2 π}{3}

\frac{\frac{4 π}{3}}{2}

Apply the fraction rule:

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

= \frac{4 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{4 π}{6}

Cancel the common factor:

2

= \frac{2 π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

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

Popular Examples

tan(x)= 10/12

tan (x) = \frac{10}{12}

0.666=sin(x)

0.666 = sin (x)

2*cos^2(x)-sin^2(x)+1=0

2 \cdot cos^{2} (x) - sin^{2} (x) + 1 = 0

1=sinh(x)

1 = sinh (x)

15cos^2(θ)+cos(θ)-2=0

15 cos^{2} (θ) + cos (θ) - 2 = 0

Frequently Asked Questions (FAQ)

What is the general solution for -1=2cos(2θ) ?
The general solution for -1=2cos(2θ) is θ= pi/3+pin,θ=(2pi)/3+pin