What is the general solution for sqrt(1-cos(3x))=2pi ?

The general solution for sqrt(1-cos(3x))=2pi is No Solution for x\in\mathbb{R}

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Popular Trigonometry >

sqrt(1-cos(3x))=2pi

Solution

1 - cos (3 x) = 2 π

Solution

No Solution for x \in R

Solution steps

1 - cos (3 x) = 2 π

Square both sides:

1 - cos (3 x) = 4 π^{2}

1 - cos (3 x) = 2 π

(1 - cos (3 x))^{2} = (2 π)^{2}

Expand

(1 - cos (3 x))^{2} : 1 - cos (3 x)

(1 - cos (3 x))^{2}

Apply radical rule:

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

= ((1 - cos (3 x))^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= (1 - cos (3 x))^{\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 - cos (3 x)

Expand

(2 π)^{2} : 4 π^{2}

(2 π)^{2}

Apply exponent rule:

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

= 2^{2} π^{2}

2^{2} = 4

= 4 π^{2}

1 - cos (3 x) = 4 π^{2}

1 - cos (3 x) = 4 π^{2}

Solve

1 - cos (3 x) = 4 π^{2} : cos (3 x) = - 4 π^{2} + 1

1 - cos (3 x) = 4 π^{2}

Move

1

to the right side

1 - cos (3 x) = 4 π^{2}

Subtract

1

from both sides

1 - cos (3 x) - 1 = 4 π^{2} - 1

Simplify

- cos (3 x) = 4 π^{2} - 1

- cos (3 x) = 4 π^{2} - 1

Divide both sides by

- 1

- cos (3 x) = 4 π^{2} - 1

Divide both sides by

- 1

\frac{- cos ( 3 x )}{- 1} = \frac{4 π ^{2}}{- 1} - \frac{1}{- 1}

Simplify

\frac{- cos ( 3 x )}{- 1} = \frac{4 π ^{2}}{- 1} - \frac{1}{- 1}

Simplify

\frac{- cos ( 3 x )}{- 1} : cos (3 x)

\frac{- cos ( 3 x )}{- 1}

Apply the fraction rule:

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

= \frac{cos ( 3 x )}{1}

Apply rule

\frac{a}{1} = a

= cos (3 x)

Simplify

\frac{4 π ^{2}}{- 1} - \frac{1}{- 1} : - 4 π^{2} + 1

\frac{4 π ^{2}}{- 1} - \frac{1}{- 1}

Apply rule

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

= \frac{4 π ^{2} - 1}{- 1}

Apply the fraction rule:

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

= - \frac{4 π ^{2} - 1}{1}

Apply rule

\frac{a}{1} = a

= - (4 π^{2} - 1)

Distribute parentheses

= - (4 π^{2}) - (- 1)

Apply minus-plus rules

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

= - 4 π^{2} + 1

cos (3 x) = - 4 π^{2} + 1

cos (3 x) = - 4 π^{2} + 1

cos (3 x) = - 4 π^{2} + 1

cos (3 x) = - 4 π^{2} + 1

Verify Solutions:

cos (3 x) = - 4 π^{2} + 1

True

Check the solutions by plugging them into

1 - cos (3 x) = 2 π

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

Plug in

cos (3 x) = - 4 π^{2} + 1 :

True

1 - (- 4 π^{2} + 1) = 2 π

1 - (- 4 π^{2} + 1) = 2 π

1 - (- 4 π^{2} + 1)

Expand

1 - (- 4 π^{2} + 1) : 4 π^{2}

1 - (- 4 π^{2} + 1)

- (- 4 π^{2} + 1) : 4 π^{2} - 1

- (- 4 π^{2} + 1)

Distribute parentheses

= - (- 4 π^{2}) - (1)

Apply minus-plus rules

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

= 4 π^{2} - 1

= 1 + 4 π^{2} - 1

1 - 1 = 0

= 4 π^{2}

= 4 π^{2}

Apply radical rule:

n ab = n a n b,

assuming

a \geq 0, b \geq 0

= 4 π^{2}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= 2 π^{2}

Apply radical rule:

n a^{n} = a,

assuming

a \geq 0

π^{2} = π

= 2 π

2 π = 2 π

True

The solution is

cos (3 x) = - 4 π^{2} + 1

- 1 \leq cos (x) \leq 1

No Solution for x \in R

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(1-cos(3x))=2pi ?
The general solution for sqrt(1-cos(3x))=2pi is No Solution for x\in\mathbb{R}