What is the general solution for sqrt((1-cos(x))/2)+cos(x)-1=0 ?

The general solution for sqrt((1-cos(x))/2)+cos(x)-1=0 is x=2pin,x= pi/3+2pin,x=(5pi)/3+2pin

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sqrt((1-cos(x))/2)+cos(x)-1=0

Solution

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

Solution

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

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

\frac{1 - u}{2} + u - 1 = 0

\frac{1 - u}{2} + u - 1 = 0 : u = 1, u = \frac{1}{2}

\frac{1 - u}{2} + u - 1 = 0

Remove square roots

\frac{1 - u}{2} + u - 1 = 0

Subtract

u

from both sides

\frac{1 - u}{2} + u - 1 - u = 0 - u

Simplify

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

Add

1

to both sides

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

Simplify

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

Square both sides:

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

\frac{1 - u}{2} + u - 1 = 0

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

Expand

(- u + 1)^{2} : u^{2} - 2 u + 1

(- u + 1)^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = - u, b = 1

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

Simplify

(- u)^{2} + 2 (- u) \cdot 1 + 1^{2} : u^{2} - 2 u + 1

(- u)^{2} + 2 (- u) \cdot 1 + 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

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

Remove parentheses:

(- a) = - a

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- u)^{2} = u^{2}

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

Multiply the numbers:

2 \cdot 1 = 2

= u^{2} - 2 u + 1

= u^{2} - 2 u + 1

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

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

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

Solve

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

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

Multiply both sides by

2

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

Multiply both sides by

2

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

Simplify

1 - u = 2 u^{2} - 4 u + 2

1 - u = 2 u^{2} - 4 u + 2

Switch sides

2 u^{2} - 4 u + 2 = 1 - u

Move

u

to the left side

2 u^{2} - 4 u + 2 = 1 - u

Add

u

to both sides

2 u^{2} - 4 u + 2 + u = 1 - u + u

Simplify

2 u^{2} - 3 u + 2 = 1

2 u^{2} - 3 u + 2 = 1

Move

1

to the left side

2 u^{2} - 3 u + 2 = 1

Subtract

1

from both sides

2 u^{2} - 3 u + 2 - 1 = 1 - 1

Simplify

2 u^{2} - 3 u + 1 = 0

2 u^{2} - 3 u + 1 = 0

Solve with the quadratic formula

2 u^{2} - 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = - 3, c = 1

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

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

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

3 + 1 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

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

Verify Solutions:

u = 1

True

, u = \frac{1}{2}

True

Check the solutions by plugging them into

\frac{1 - u}{2} + u - 1 = 0

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

Plug in

u = 1 :

True

\frac{1 - 1}{2} + 1 - 1 = 0

\frac{1 - 1}{2} + 1 - 1 = 0

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

\frac{1 - 1}{2} = 0

\frac{1 - 1}{2}

\frac{1 - 1}{2} = 0

\frac{1 - 1}{2}

Subtract the numbers:

1 - 1 = 0

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

= 0

Apply rule

0 = 0

= 0

= 0 + 1 - 1

Add/Subtract the numbers:

0 + 1 - 1 = 0

= 0

0 = 0

True

Plug in

u = \frac{1}{2} :

True

\frac{1 - ( \frac{1}{2} )}{2} + (\frac{1}{2}) - 1 = 0

\frac{1 - ( \frac{1}{2} )}{2} + (\frac{1}{2}) - 1 = 0

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

Remove parentheses:

(a) = a

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

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

\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:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

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

Combine the fractions

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

Apply rule

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

= \frac{1 + 1}{2}

Add the numbers:

1 + 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

= 1 - 1

Subtract the numbers:

1 - 1 = 0

= 0

0 = 0

True

The solutions are

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

Substitute back

u = cos (x)

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

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

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 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}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt((1-cos(x))/2)+cos(x)-1=0 ?
The general solution for sqrt((1-cos(x))/2)+cos(x)-1=0 is x=2pin,x= pi/3+2pin,x=(5pi)/3+2pin