What is the general solution for cos(x)=sqrt((1-cos(x))/2) ?

The general solution for cos(x)=sqrt((1-cos(x))/2) is x= pi/3+2pin,x=(5pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(x)=sqrt((1-cos(x))/2)

Solution

cos (x) = \frac{1 - cos ( x )}{2}

Solution

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

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (x) = \frac{1 - cos ( x )}{2}

Solve by substitution

cos (x) = \frac{1 - cos ( x )}{2}

Let:

cos (x) = u

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

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

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

Square both sides:

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

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

u^{2} = (\frac{1 - u}{2})^{2}

Expand

(\frac{1 - u}{2})^{2} : \frac{1 - u}{2}

(\frac{1 - u}{2})^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((\frac{1 - u}{2})^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (\frac{1 - u}{2})^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= \frac{1 - u}{2}

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

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

Solve

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

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

Multiply both sides by

2

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

Multiply both sides by

2

u^{2} \cdot 2 = \frac{1 - u}{2} \cdot 2

Simplify

2 u^{2} = 1 - u

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

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

Verify Solutions:

u = \frac{1}{2}

True

, u = - 1

False

Check the solutions by plugging them into

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

Remove the ones that don't agree with the equation.

Plug in

u = \frac{1}{2} :

True

\frac{1}{2} = \frac{1 - ( \frac{1}{2} )}{2}

\frac{1 - ( \frac{1}{2} )}{2} = \frac{1}{2}

\frac{1 - ( \frac{1}{2} )}{2}

Remove parentheses:

(a) = a

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

\frac{1 - \frac{1}{2}}{2} = \frac{1}{4}

\frac{1 - \frac{1}{2}}{2}

Join

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

1 - \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 2 - 1 = 1

1 \cdot 2 - 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= \frac{1}{2}

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

Apply the fraction rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1}{4}

= \frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

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

True

Plug in

u = - 1 :

False

- 1 = \frac{1 - ( - 1 )}{2}

\frac{1 - ( - 1 )}{2} = 1

\frac{1 - ( - 1 )}{2}

Apply rule

- (- a) = a

= \frac{1 + 1}{2}

\frac{1 + 1}{2} = 1

\frac{1 + 1}{2}

Add the numbers:

1 + 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

= 1

Apply rule

1 = 1

= 1

- 1 = 1

False

The solution is

u = \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

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

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \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}

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

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

Combine all the solutions

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

Popular Examples

7cos(4x)=6

7 cos (4 x) = 6

sin(x)tan(x)+3tan(x)=0

sin (x) tan (x) + 3 tan (x) = 0

sin^2(x/2)=0

sin^{2} (\frac{x}{2}) = 0

tan(t)=2

tan (t) = 2

-120=-90-arctan(0.1)+arctan(0.1x)

- 120 = - 90 - arctan (0.1) + arctan (0.1 x)

Frequently Asked Questions (FAQ)

What is the general solution for cos(x)=sqrt((1-cos(x))/2) ?
The general solution for cos(x)=sqrt((1-cos(x))/2) is x= pi/3+2pin,x=(5pi)/3+2pin