What is the general solution for sqrt(4cos(θ))=1-cos(θ) ?

The general solution for sqrt(4cos(θ))=1-cos(θ) is θ=1.39837…+2pin,θ=2pi-1.39837…+2pin

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sqrt(4cos(θ))=1-cos(θ)

Solution

4 cos (θ) = 1 - cos (θ)

Solution

θ = 1.39837 \dots + 2 πn, θ = 2 π - 1.39837 \dots + 2 πn

Degrees

θ = 80.12071 \dots^{\circ} + 36 0^{\circ} n, θ = 279.87928 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (θ) = 1 - cos (θ)

Solve by substitution

4 cos (θ) = 1 - cos (θ)

Let:

cos (θ) = u

4 u = 1 - u

4 u = 1 - u : u = 3 - 22

4 u = 1 - u

Square both sides:

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

4 u = 1 - u

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

Expand

(4 u)^{2} : 4 u

(4 u)^{2}

Apply radical rule:

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

= ((4 u)^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= (4 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

= 4 u

Expand

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

(1 - u)^{2}

Apply Perfect Square Formula:

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

a = 1, b = u

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

Multiply the numbers:

2 \cdot 1 = 2

= 1 - 2 u + u^{2}

= 1 - 2 u + u^{2}

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

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

Solve

4 u = 1 - 2 u + u^{2} : u = 3 + 22, u = 3 - 22

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

Switch sides

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

Move

4 u

to the left side

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

Subtract

4 u

from both sides

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

Simplify

u^{2} - 6 u + 1 = 0

u^{2} - 6 u + 1 = 0

Solve with the quadratic formula

u^{2} - 6 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 6, c = 1

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 1 = 42

(- 6)^{2} - 4 \cdot 1 \cdot 1

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 6^{2} - 4

6^{2} = 36

= 36 - 4

Subtract the numbers:

36 - 4 = 32

= 32

Prime factorization of

32 : 2^{5}

32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

Apply exponent rule:

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

= 2^{4} \cdot 2

Apply radical rule:

= 2 2^{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2

Refine

= 42

u_{1, 2} = \frac{- ( - 6 ) \pm 4 2}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 4 2}{2 \cdot 1}, u_{2} = \frac{- ( - 6 ) - 4 2}{2 \cdot 1}

u = \frac{- ( - 6 ) + 4 2}{2 \cdot 1} : 3 + 22

\frac{- ( - 6 ) + 4 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 + 4 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6 + 4 2}{2}

Factor

6 + 42 : 2 (3 + 22)

6 + 42

Rewrite as

= 2 \cdot 3 + 2 \cdot 22

Factor out common term

2

= 2 (3 + 22)

= \frac{2 ( 3 + 2 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 3 + 22

u = \frac{- ( - 6 ) - 4 2}{2 \cdot 1} : 3 - 22

\frac{- ( - 6 ) - 4 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 - 4 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6 - 4 2}{2}

Factor

6 - 42 : 2 (3 - 22)

6 - 42

Rewrite as

= 2 \cdot 3 - 2 \cdot 22

Factor out common term

2

= 2 (3 - 22)

= \frac{2 ( 3 - 2 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 3 - 22

The solutions to the quadratic equation are:

u = 3 + 22, u = 3 - 22

u = 3 + 22, u = 3 - 22

Verify Solutions:

u = 3 + 22

False

, u = 3 - 22

True

Check the solutions by plugging them into

4 u = 1 - u

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

Plug in

u = 3 + 22 :

False

4 (3 + 22) = 1 - (3 + 22)

1 - (3 + 22) = - 2 - 22

1 - (3 + 22)

- (3 + 22) : - 3 - 22

- (3 + 22)

Distribute parentheses

= - (3) - (22)

Apply minus-plus rules

+ (- a) = - a

= - 3 - 22

= 1 - 3 - 22

Subtract the numbers:

1 - 3 = - 2

= - 2 - 22

4 (3 + 22) = - 2 - 22

False

Plug in

u = 3 - 22 :

True

4 (3 - 22) = 1 - (3 - 22)

1 - (3 - 22) = 22 - 2

1 - (3 - 22)

- (3 - 22) : - 3 + 22

- (3 - 22)

Distribute parentheses

= - (3) - (- 22)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 3 + 22

= 1 - 3 + 22

Subtract the numbers:

1 - 3 = - 2

= 22 - 2

4 (3 - 22) = 22 - 2

True

The solution is

u = 3 - 22

Substitute back

u = cos (θ)

cos (θ) = 3 - 22

cos (θ) = 3 - 22

cos (θ) = 3 - 22 : θ = arccos (3 - 22) + 2 πn, θ = 2 π - arccos (3 - 22) + 2 πn

cos (θ) = 3 - 22

Apply trig inverse properties

cos (θ) = 3 - 22

General solutions for

cos (θ) = 3 - 22

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

θ = arccos (3 - 22) + 2 πn, θ = 2 π - arccos (3 - 22) + 2 πn

θ = arccos (3 - 22) + 2 πn, θ = 2 π - arccos (3 - 22) + 2 πn

Combine all the solutions

θ = arccos (3 - 22) + 2 πn, θ = 2 π - arccos (3 - 22) + 2 πn

Show solutions in decimal form

θ = 1.39837 \dots + 2 πn, θ = 2 π - 1.39837 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(4cos(θ))=1-cos(θ) ?
The general solution for sqrt(4cos(θ))=1-cos(θ) is θ=1.39837…+2pin,θ=2pi-1.39837…+2pin