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=0.87579…+2pin,x=2pi-0.87579…+2pin

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

Solution

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

Solution

x = 0.87579 \dots + 2 πn, x = 2 π - 0.87579 \dots + 2 πn

Degrees

x = 50.17922 \dots^{\circ} + 36 0^{\circ} n, x = 309.82077 \dots^{\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 + 17}{8}

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

Multiply both sides by

2

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

Simplify

u \cdot 2 = 1 + u

Square both sides:

4 u^{2} = 1 + u

u \cdot 2 = 1 + u

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

Expand

(u \cdot 2)^{2} : 4 u^{2}

(u \cdot 2)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} u^{2}

2^{2} = 4

= 4 u^{2}

Expand

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

(1 + u)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= (1 + u)^{\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

= 1 + u

4 u^{2} = 1 + u

4 u^{2} = 1 + u

Solve

4 u^{2} = 1 + u : u = \frac{1 + 17}{8}, u = \frac{1 - 17}{8}

4 u^{2} = 1 + u

Move

u

to the left side

4 u^{2} = 1 + u

Subtract

u

from both sides

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

Simplify

4 u^{2} - u = 1

4 u^{2} - u = 1

Move

1

to the left side

4 u^{2} - u = 1

Subtract

1

from both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = - 1, c = - 1

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

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

(- 1)^{2} - 4 \cdot 4 (- 1) = 17

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

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 4 \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 4 \cdot 1 = 16

4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 16

= 1 + 16

Add the numbers:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 17}{2 \cdot 4}, u_{2} = \frac{- ( - 1 ) - 17}{2 \cdot 4}

u = \frac{- ( - 1 ) + 17}{2 \cdot 4} : \frac{1 + 17}{8}

\frac{- ( - 1 ) + 17}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{1 + 17}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{1 + 17}{8}

u = \frac{- ( - 1 ) - 17}{2 \cdot 4} : \frac{1 - 17}{8}

\frac{- ( - 1 ) - 17}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{1 - 17}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{1 - 17}{8}

The solutions to the quadratic equation are:

u = \frac{1 + 17}{8}, u = \frac{1 - 17}{8}

u = \frac{1 + 17}{8}, u = \frac{1 - 17}{8}

Verify Solutions:

u = \frac{1 + 17}{8}

True

, u = \frac{1 - 17}{8}

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 + 17}{8} :

True

\frac{1 + 17}{8} = \frac{1 + ( \frac{1 + 17}{8} )}{2}

\frac{1 + ( \frac{1 + 17}{8} )}{2} = \frac{9 + 17}{4 2}

\frac{1 + ( \frac{1 + 17}{8} )}{2}

Remove parentheses:

(a) = a

= \frac{1 + \frac{1 + 17}{8}}{2}

1 + \frac{1 + 17}{8} = \frac{9 + 17}{2 2}

1 + \frac{1 + 17}{8}

Join

1 + \frac{1 + 17}{8} : \frac{9 + 17}{8}

1 + \frac{1 + 17}{8}

Convert element to fraction:

1 = \frac{1 \cdot 8}{8}

= \frac{1 \cdot 8}{8} + \frac{1 + 17}{8}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 8 + 1 + 17}{8}

1 \cdot 8 + 1 + 17 = 9 + 17

1 \cdot 8 + 1 + 17

Multiply the numbers:

1 \cdot 8 = 8

= 8 + 1 + 17

Add the numbers:

8 + 1 = 9

= 9 + 17

= \frac{9 + 17}{8}

= \frac{9 + 17}{8}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{9 + 17}{8}

8 = 22

8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

= \frac{9 + 17}{2 2}

= \frac{\frac{9 + 17}{2 2}}{2}

Apply the fraction rule:

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

= \frac{9 + 17}{2 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{9 + 17}{4 2}

\frac{1 + 17}{8} = \frac{9 + 17}{4 2}

True

Plug in

u = \frac{1 - 17}{8} :

False

\frac{1 - 17}{8} = \frac{1 + ( \frac{1 - 17}{8} )}{2}

\frac{1 + ( \frac{1 - 17}{8} )}{2} = \frac{9 - 17}{4 2}

\frac{1 + ( \frac{1 - 17}{8} )}{2}

Remove parentheses:

(a) = a

= \frac{1 + \frac{1 - 17}{8}}{2}

1 + \frac{1 - 17}{8} = \frac{9 - 17}{2 2}

1 + \frac{1 - 17}{8}

Join

1 + \frac{1 - 17}{8} : \frac{9 - 17}{8}

1 + \frac{1 - 17}{8}

Convert element to fraction:

1 = \frac{1 \cdot 8}{8}

= \frac{1 \cdot 8}{8} + \frac{1 - 17}{8}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 8 + 1 - 17}{8}

1 \cdot 8 + 1 - 17 = 9 - 17

1 \cdot 8 + 1 - 17

Multiply the numbers:

1 \cdot 8 = 8

= 8 + 1 - 17

Add the numbers:

8 + 1 = 9

= 9 - 17

= \frac{9 - 17}{8}

= \frac{9 - 17}{8}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{9 - 17}{8}

8 = 22

8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

= \frac{9 - 17}{2 2}

= \frac{\frac{9 - 17}{2 2}}{2}

Apply the fraction rule:

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

= \frac{9 - 17}{2 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{9 - 17}{4 2}

\frac{1 - 17}{8} = \frac{9 - 17}{4 2}

False

The solution is

u = \frac{1 + 17}{8}

Substitute back

u = cos (x)

cos (x) = \frac{1 + 17}{8}

cos (x) = \frac{1 + 17}{8}

cos (x) = \frac{1 + 17}{8} : x = arccos (\frac{1 + 17}{8}) + 2 πn, x = 2 π - arccos (\frac{1 + 17}{8}) + 2 πn

cos (x) = \frac{1 + 17}{8}

Apply trig inverse properties

cos (x) = \frac{1 + 17}{8}

General solutions for

cos (x) = \frac{1 + 17}{8}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1 + 17}{8}) + 2 πn, x = 2 π - arccos (\frac{1 + 17}{8}) + 2 πn

x = arccos (\frac{1 + 17}{8}) + 2 πn, x = 2 π - arccos (\frac{1 + 17}{8}) + 2 πn

Combine all the solutions

x = arccos (\frac{1 + 17}{8}) + 2 πn, x = 2 π - arccos (\frac{1 + 17}{8}) + 2 πn

Show solutions in decimal form

x = 0.87579 \dots + 2 πn, x = 2 π - 0.87579 \dots + 2 πn

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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=0.87579…+2pin,x=2pi-0.87579…+2pin