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

The general solution for sqrt(cos(θ))=2cos(θ)-1 is θ=2pin

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sqrt(cos(θ))=2cos(θ)-1

Solution

cos (θ) = 2 cos (θ) - 1

Solution

θ = 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (θ) = 2 cos (θ) - 1

Solve by substitution

cos (θ) = 2 cos (θ) - 1

Let:

cos (θ) = u

u = 2 u - 1

u = 2 u - 1 : u = 1

u = 2 u - 1

Square both sides:

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

u = 2 u - 1

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

Expand

(u)^{2} : u

(u)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 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

= u

Expand

(2 u - 1)^{2} : 4 u^{2} - 4 u + 1

(2 u - 1)^{2}

Apply Perfect Square Formula:

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

a = 2 u, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

(2 u)^{2}

Apply exponent rule:

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

= 2^{2} u^{2}

2^{2} = 4

= 4 u^{2}

2 \cdot 2 u \cdot 1 = 4 u

2 \cdot 2 u \cdot 1

Multiply the numbers:

2 \cdot 2 \cdot 1 = 4

= 4 u

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

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

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

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

Solve

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

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

Switch sides

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

Move

u

to the left side

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

Subtract

u

from both sides

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

Simplify

4 u^{2} - 5 u + 1 = 0

4 u^{2} - 5 u + 1 = 0

Solve with the quadratic formula

4 u^{2} - 5 u + 1 = 0

Quadratic Equation Formula:

For

a = 4, b = - 5, c = 1

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

(- 5)^{2} - 4 \cdot 4 \cdot 1 = 3

(- 5)^{2} - 4 \cdot 4 \cdot 1

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 4}, u_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 4}

u = \frac{- ( - 5 ) + 3}{2 \cdot 4} : 1

\frac{- ( - 5 ) + 3}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{5 + 3}{2 \cdot 4}

Add the numbers:

5 + 3 = 8

= \frac{8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 5 ) - 3}{2 \cdot 4} : \frac{1}{4}

\frac{- ( - 5 ) - 3}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{5 - 3}{2 \cdot 4}

Subtract the numbers:

5 - 3 = 2

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2}{8}

Cancel the common factor:

2

= \frac{1}{4}

The solutions to the quadratic equation are:

u = 1, u = \frac{1}{4}

u = 1, u = \frac{1}{4}

Verify Solutions:

u = 1

True

, u = \frac{1}{4}

False

Check the solutions by plugging them into

u = 2 u - 1

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

Plug in

u = 1 :

True

1 = 2 \cdot 1 - 1

1 = 1

1

Apply rule

1 = 1

= 1

2 \cdot 1 - 1 = 1

2 \cdot 1 - 1

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

1 = 1

True

Plug in

u = \frac{1}{4} :

False

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

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

\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}

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

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

Remove parentheses:

(a) = a

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

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

2 \cdot \frac{1}{4}

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

= \frac{1}{2} - 1

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

Add/Subtract the numbers:

- 2 + 1 = - 1

= - 1

= \frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

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

False

The solution is

u = 1

Substitute back

u = cos (θ)

cos (θ) = 1

cos (θ) = 1

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

General solutions for

cos (θ) = 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}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Combine all the solutions

θ = 2 πn

Popular Examples

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2 cos^{2} (2 β) + cos (2 β) - 1 = 0, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

cos(2x)= 3/7

cos (2 x) = \frac{3}{7}

0=1-3cos(θ)

0 = 1 - 3 cos (θ)

(tan((3a)/2))tan(a/2)=3

(tan (\frac{3 a}{2})) tan (\frac{a}{2}) = 3

1=sqrt(3)sin(x)

1 = 3 sin (x)

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(cos(θ))=2cos(θ)-1 ?
The general solution for sqrt(cos(θ))=2cos(θ)-1 is θ=2pin