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=2pin

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

Solution

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

Solution

x = 2 πn

Degrees

x = 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 = 1

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

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

Subtract

u

from 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

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 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} : 1

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

Apply rule

- (- a) = a

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

Add the numbers:

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 3 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

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

= - \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}

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 = 1 :

True

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

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

\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

True

Plug in

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

False

- \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}

False

The solution is

u = 1

Substitute back

u = cos (x)

cos (x) = 1

cos (x) = 1

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

Combine all the solutions

x = 2 πn

Popular Examples

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

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

sin(x/2)+(sqrt(3))/2 =0

sin (\frac{x}{2}) + \frac{3}{2} = 0

cos(x)=0.818

cos (x) = 0.818

sin(x)=(150)/(212.6)(sin(115))

sin (x) = \frac{150}{212.6} (sin (11 5^{\circ}))

16cos(2θ)-9=0

16 cos (2 θ) - 9 = 0

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=2pin