What is the general solution for cos^2(x)=cos(x)-1 ?

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

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cos^2(x)=cos(x)-1

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

u^{2} = u - 1

u^{2} = u - 1 : u = \frac{1}{2} + i \frac{3}{2}, u = \frac{1}{2} - i \frac{3}{2}

u^{2} = u - 1

Move

1

to the left side

u^{2} = u - 1

Add

1

to both sides

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

Simplify

u^{2} + 1 = u

u^{2} + 1 = u

Move

u

to the left side

u^{2} + 1 = u

Subtract

u

from both sides

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

Simplify

u^{2} + 1 - u = 0

u^{2} + 1 - u = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} - u + 1 = 0

Solve with the quadratic formula

u^{2} - u + 1 = 0

Quadratic Equation Formula:

For

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

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

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

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 1 : 3 i

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

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 - 4

Subtract the numbers:

1 - 4 = - 3

= - 3

Apply radical rule:

- a = - 1 a

- 3 = - 1 3

= - 1 3

Apply imaginary number rule:

- 1 = i

= 3 i

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Rewrite

\frac{1 + 3 i}{2}

in standard complex form:

\frac{1}{2} + \frac{3}{2} i

\frac{1 + 3 i}{2}

Apply the fraction rule:

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

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

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Rewrite

\frac{1 - 3 i}{2}

in standard complex form:

\frac{1}{2} - \frac{3}{2} i

\frac{1 - 3 i}{2}

Apply the fraction rule:

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

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

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

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

The solutions to the quadratic equation are:

u = \frac{1}{2} + i \frac{3}{2}, u = \frac{1}{2} - i \frac{3}{2}

Substitute back

u = cos (x)

cos (x) = \frac{1}{2} + i \frac{3}{2}, cos (x) = \frac{1}{2} - i \frac{3}{2}

cos (x) = \frac{1}{2} + i \frac{3}{2}, cos (x) = \frac{1}{2} - i \frac{3}{2}

cos (x) = \frac{1}{2} + i \frac{3}{2} :

No Solution

cos (x) = \frac{1}{2} + i \frac{3}{2}

No Solution

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

No Solution

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

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

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